• 整理各种模板(准备随时弃坑)


    线段树1(真题

    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<cstdio> 
    using namespace std;
    const int N=1e5+100;
    long long tr[N*4],flag[N*4],ans;
    void built(int i,int l,int r,int x,long long val)
    {
        if(l==r)
        {
            tr[i]+=val; 
            return;
        }
        int mid=(l+r)/2;
        if(x<=mid) built(i*2,l,mid,x,val); 
        if(x>mid) built(i*2+1,mid+1,r,x,val); 
        tr[i]=tr[i*2]+tr[i*2+1]; 
        return;
    }
    void pushdown(int i,int l,int r) 
    {
        int mid=(l+r)/2;
        tr[i*2]+=flag[i]*(mid-l+1);
        tr[i*2+1]+=flag[i]*(r-mid); 
        flag[i*2]+=flag[i];
        flag[i*2+1]+=flag[i];
        flag[i]=0; 
        return;
    }
    void update(int i,int l,int r,long long x,long long y,long long k) 
    {
        if(l>=x&&r<=y)
        {
            tr[i]+=k*(r-l+1);
            flag[i]+=k; 
            return;
        }
        if(flag[i]>0) pushdown(i,l,r);
        int mid=(l+r)/2;
        if(mid>=x) update(i*2,l,mid,x,y,k); 
        if(mid<y) update(i*2+1,mid+1,r,x,y,k);
        tr[i]=tr[i*2]+tr[i*2+1];
        return;
    }
    void query(int i,int l,int r,long long x,long long y)
    {
        if(l>=x&&r<=y)
        {
            ans+=tr[i]; 
            return;
        }
        if(flag[i]>0) pushdown(i,l,r); 
        int mid=(l+r)/2;
        if(mid>=x) query(i*2,l,mid,x,y); 
        if(mid<y) query(i*2+1,mid+1,r,x,y); 
    }
    int main()
    {
        int n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        {
            long long x;
            cin>>x;
            built(1,1,n,i,x);
        }
        while(m--)
        {
            int op;
            cin>>op;
            if(op==1)
            {
                long long x,y,k;
                cin>>x>>y>>k;
                update(1,1,n,x,y,k);
            }
            if(op==2)
            {
                long long x,y;
                cin>>x>>y;
                ans=0;
                query(1,1,n,x,y);
                cout<<ans<<endl;
            }
        }
        return 0;
    }
    

    线段树2(真题

    #include<bits/stdc++.h>
    using namespace std;
    const int N = 1e5+100;
    int n,m;
    int a,b,c;
    long long sum[N*4],fa[N*4],fm[N*4],w,p,ans;
    bool mu[N*4];
    int mid[N*4];
    void build(int i,int l,int r)
    {
        fm[i]=1;
        if(l==r)
        {
            cin>>sum[i];
            return;
        }
        mid[i]=(l+r)/2;
        build(i*2,l,mid[i]);
        build(i*2+1,mid[i]+1,r);
        sum[i]=(sum[i*2]+sum[i*2+1])%p;
    }
    void down(int i,int l,int mid,int r)
    {
        if(mu[i])
        {
            sum[i*2]=sum[i*2]*fm[i]%p;
            sum[i*2+1]=sum[i*2+1]*fm[i]%p;
            fa[i*2]=fa[i*2]*fm[i]%p;
            fa[i*2+1]=fa[i*2+1]*fm[i]%p;
            fm[i*2]=fm[i*2]*fm[i]%p;
            fm[i*2+1]=fm[i*2+1]*fm[i]%p;
            mu[i*2]=mu[i*2+1]=true;
            fm[i]=1;
            mu[i]=false;	
        }
        if(fa[i])
        {
            sum[i*2]=(sum[i*2]+fa[i]*(mid-l+1)%p)%p;
            sum[i*2+1]=(sum[i*2+1]+fa[i]*(r-mid)%p)%p;
            fa[i*2]=(fa[i*2]+fa[i])%p;
            fa[i*2+1]=(fa[i*2+1]+fa[i])%p;
            fa[i]=0;
        }
    }
    void solve(int i,int l,int r)
    {
        if(l>=b && r<=c)
        {
            if(a==1)
            {
                sum[i]=sum[i]*w%p;
                fa[i]=fa[i]*w%p;
                fm[i]=fm[i]*w%p;
                mu[i]=true;
            }
            else
            if(a==2)
            {
                sum[i]=(sum[i]+w*(r-l+1)%p)%p;
                fa[i]+=w;
            }
            else
            ans=(ans+sum[i])%p;
            return;
        }
        if(mu[i] || fa[i]) down(i,l,mid[i],r);
        if(b<=mid[i]) solve(i*2,l,mid[i]);
        if(c>mid[i]) solve(i*2+1,mid[i]+1,r);
        sum[i]=(sum[i*2]+sum[i*2+1])%p;
    }
    int main()
    {
        cin>>n>>m>>p;
        build(1,1,n);
        while(m--)
        {
            cin>>a;
            if(a==1 || a==2)
            {
                cin>>b>>c>>w;
                solve(1,1,n);
            }
            else
            {
                ans=0;
                cin>>b>>c;
                solve(1,1,n);
                cout<<ans<<endl;
            }
        }
        return 0;
    }
    

    ST表(真题

    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    inline int read()
    {
        char c=getchar();
        int x=0,f=1;
        while(c<'0'||c>'9')
        {
            if(c=='-')f=-1;
            c=getchar();
        }
        while(c>='0'&&c<='9')
        {
            x=x*10+c-'0';
            c=getchar();
        }
        return x*f;
    }
    int maxx[1000010][21];
    int query(int l,int r)
    {
        int k=log2(r-l+1);
        return max(maxx[l][k],maxx[r-(1<<k)+1][k]);
    }
    int main()
    {
        int n=read(),m=read();
        for(int i=1; i<=n; i++)
            maxx[i][0]=read();
        for(int j=1; j<=21; j++)
            for(int i=1; i+(1<<j)-1<=n; i++)
                maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);
        for(int i=1; i<=m; i++)
        {
            int l=read(),r=read();
            printf("%d
    ",query(l,r));
        }
        return 0;
    }
    

    并查集(真题

    #include<bits/stdc++.h>
    using namespace std;
    int n,m;
    int x,y,z;
    int a[11000];
    int find(int x)
    {
        if(x==a[x])
            return x;
        else
        return a[x]=find(a[x]);
    }
    int main()
    {
        cin>>n>>m;
        while(n--)
            a[n]=n;
        while(m--)
        {
            cin>>x>>y>>z;
            if(x&2)
            {
                if(find(y)==find(z))
                    cout<<"Y"<<endl;
                else
                    cout<<"N"<<endl;
            }
            else
            if(find(y)!=find(z))
                a[find(y)]=find(z);
        }
        return 0;
    }
    

    乘法逆元(真题

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cmath>
    #define int long long
    using namespace std;
    int a[3000010];
    int n,p;
    main()
    {
        cin>>n>>p;
        a[1]=1;
        cout<<1<<"
    ";
        for(int i=2;i<=n;i++)
        {
            a[i]=(p-p/i)*a[p%i]%p;
            cout<<a[i]<<"
    ";
        }
        return 0;
    }
    

    单源最短路(真题

    #include <bits/stdc++.h>
    #define re register
    using namespace std;
    inline int read() {
        int X=0,w=1; char c=getchar();
        while (c<'0'||c>'9') { if (c=='-') w=-1; c=getchar(); }
        while (c>='0'&&c<='9') X=(X<<3)+(X<<1)+c-'0',c=getchar();
        return X*w;
    }
    struct Edge { int v,w,nxt; };
    Edge e[500010];
    int head[100010],cnt=0;
    inline void addEdge(int u,int v,int w) {
        e[++cnt].v=v;
        e[cnt].w=w;
        e[cnt].nxt=head[u];
        head[u]=cnt;
    }
    int n,m,s;
    int dis[100010];
    struct node { //堆节点
        int u,d;
        bool operator <(const node& rhs) const {
            return d>rhs.d;
        }
    };
    inline void Dijkstra() {
        for (re int i=1;i<=n;i++) dis[i]=2147483647;
        dis[s]=0;
        priority_queue<node> Q; //堆
        Q.push((node){s,0});
        while (!Q.empty()) {
            node fr=Q.top(); Q.pop();
            int u=fr.u,d=fr.d;
            if (d!=dis[u]) continue;
            for (re int i=head[u];i;i=e[i].nxt) {
                int v=e[i].v,w=e[i].w;
                if (dis[u]+w<dis[v]) {
                    dis[v]=dis[u]+w;
                    Q.push((node){v,dis[v]});
                }
            }
        }
    }
    int main() {
        n=read(),m=read(),s=read();
        for (re int i=1;i<=m;i++) {
            int X=read(),Y=read(),Z=read();
            addEdge(X,Y,Z);
        }
        Dijkstra();
        for (re int i=1;i<=n;i++) printf("%d ",dis[i]);
        return 0;
    }
    

    堆(真题)

    #include<bits/stdc++.h>
    using namespace std;
    priority_queue< int,vector<int>,greater<int> >a;
    int n,m;
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&m);
            if(m==1)
            {
                scanf("%d",&m);
                a.push(m);
            }
            else
            if(m==2)
            {
                cout<<a.top()<<endl;
            }
            else
            a.pop();
        }
        return 0;
    }
    

    二分图匹配(真题

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    using namespace std;
    int map[1010][1010];
    int pl[1000010];
    bool mark[1000010];
    int n,m,e,tot;
    inline int read()
    {
        int x=0,f=1;
        char ch=getchar();
        for(; !isdigit(ch); ch=getchar())
            if(ch=='-')
                f=-1;
        for(; isdigit(ch); ch=getchar())
            x=(x<<1)+(x<<3)+ch-'0';
        return x*f;
    }
    bool find(int x)
    {
        for(int i=1; i<=m; i++)
        {
            if(map[x][i]==1&&mark[i]==0)
            {
                mark[i]=1;
                if(pl[i]==0||find(pl[i])==1)
                {
                    pl[i]=x;
                    return 1;
                }
            }
        }
        return 0;
    }
    int main()
    {
        n=read(),m=read(),e=read();
        for(int i=1; i<=e; i++)
        {
            int x=read(),y=read();
            if(x<=n&&y<=m)
                map[x][y]=1;
        }
        for(int i=1; i<=n; i++)
        {
            memset(mark,0,sizeof(mark));
            if(find(i)==1)
                tot++;
        }
        cout<<tot;
        return 0;
    }
    

    负环(真题

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<queue>
    using namespace std;
    
    typedef long long LL;
    
    inline int read()
    {
        char c=getchar();int num=0,f=1;
        for(;!isdigit(c);c=getchar())
            f=c=='-'?-1:f;
        for(;isdigit(c);c=getchar())
            num=num*10+c-'0';
        return num*f;
    }
    
    const int N=2e3+5;
    const int M=3e3+5;
    
    int T,n,m;
    
    int head[N],num_edge;
    struct Edge
    {
        int v,w,nxt;
    }edge[M<<1];
    
    inline void add_edge(int u,int v,int w)
    {
        edge[++num_edge].v=v;
        edge[num_edge].w=w;
        edge[num_edge].nxt=head[u];
        head[u]=num_edge;
    }
    
    int dis[N],cnt[N],vis[N],tim;
    queue<int> que;
    bool spfa()
    {
        ++tim;
        memset(dis,0x3f,sizeof(dis));
        memset(cnt,0,sizeof(cnt));
        while(!que.empty()) que.pop();
        int u;dis[1]=0,cnt[1]=1;que.push(1);
        while(!que.empty())
        {
            u=que.front(),que.pop();
            vis[u]=0;
            for(int i=head[u],v;i;i=edge[i].nxt)
            {
                v=edge[i].v;
                if(dis[v]>dis[u]+edge[i].w)
                {
                    dis[v]=dis[u]+edge[i].w;
                    ++cnt[v]/*=cnt[u]+1*/;
                    if(cnt[v]>=n) return 1;
                    if(vis[v]!=tim)
                    {
                        vis[v]=tim;
                        que.push(v);
                    }
                }
            }
        }
        return 0;
    }
    
    int main()
    {
        T=read();
        while(T--)
        {
            memset(head,0,sizeof(head));num_edge=0;
            n=read(),m=read();
            for(int i=1,a,b,c;i<=m;++i)
            {
                a=read(),b=read(),c=read();
                add_edge(a,b,c);
                if(c>=0) add_edge(b,a,c);
            }
            if(spfa()) puts("YE5");
            else puts("N0");
        }
        return 0;
    }
    

    矩阵快速幂(真题

    #include<iostream>
    #include<cstring>
    #define mod 1000000007
    #define ll long long
    using namespace std;
    struct Mat{
        ll m[101][101];
    };
    Mat a,e;
    ll n,p;
    Mat Mul(Mat x,Mat y) 
    {
        Mat c;
        for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
            c.m[i][j]=0;
        for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
            for(int k=1;k<=n;k++)
            {
                c.m[i][j]=c.m[i][j]%mod+x.m[i][k]*y.m[k][j]%mod;
              }
        return c; 
    }
    Mat pow(Mat x,ll y)
    {
        Mat ans=e;
        while(y)
        {
            if(y&1)
             ans=Mul(ans,x);  
            x=Mul(x,x);
            y>>=1;
        }
        return ans;
    }
    
    int main()
    {
        cin>>n>>p;
        for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
            cin>>a.m[i][j];
        for(int i=1;i<=n;i++)
            e.m[i][i]=1;    
        Mat ans=pow(a,p);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
              cout<<ans.m[i][j]%mod<<" ";
            cout<<endl;
        }  
        return 0;
    }
    

    缩点(真题

    #include<iostream>
    using namespace std;
    const int maxn = 10007;
    const int maxm = 100007;  
    struct node
    {
        int x,y;
    }E[maxm],ne[maxm];
    int n,m,tm = 0,top = 0,tot = 0;
    int W[maxn],dfn[maxn],low[maxn],Head[maxn],Next[maxm],S[maxn],V[maxn],sd[maxn];//w 点权...sd 缩点后 旧点所对应的新点
    int nh[maxn],nn[maxm],in[maxn],D[maxn];//newhead,newnext,index入度,D就是上面说的F啦
    void tarjan(int x)
    {
        dfn[x] = low[x] = ++tm;
        S[++top] = x; V[x] = 1;
        for (int y,i = Head[x]; i; i = Next[i])
        {
            y = E[i].y;
            if (!dfn[y])
            {
                tarjan(y);
                low[x] = min(low[x],low[y]);
            }
            else if (V[y])
            {
                low[x] = min(low[x],low[y]);// low[x] = min(low[x],dfn[y]);两种都行,老师讲的注释里的 求大佬解释下到底哪种好
            }
        }
        if (dfn[x] == low[x])
        {
            int y;
            while (y = S[top--])
            {
                  V[y] = 0;
                  sd[y] = x;//缩点
                  if (y==x) break;
                  W[x] +=W[y];
            }
        }
        return;
    }
    int main()
    {
        cin>>n>>m;
        for (int i = 1; i<=n; i++)
        {
            scanf("%d",&W[i]);
        }
        for (int i = 1; i<=m; i++)
        {
            scanf("%d%d",&E[i].x,&E[i].y);
            Next[i] = Head[E[i].x];
            Head[E[i].x] = i;
        }    
        for (int i = 1; i<=n; i++)
        {
            if (!dfn[i])
            {
                tarjan(i);
            }
        }
        for (int x,y,i = 1; i<=m; i++)
        {
            x = sd[E[i].x]; y = sd[E[i].y];
            if (x!=y)//不是一个强连通分量 说明需要建边
            {
                ne[++tot].x = x; ne[tot].y = y;
                nn[tot] = nh[x];
                nh[x] = tot;
                in[y]++;
            }
        }
        top =  0;
        for (int i = 1; i<=n; i++)
        {
            if (sd[i] == i && !in[i])//找那些入度为零的强连通分量根节点
            {
                S[++top] = i;
                D[i] = W[i];
            }
        }
        while (top)
        {
            int x = S[top--];
            for (int y,i = nh[x]; i; i = nn[i])
            {
                y = ne[i].y;
                D[y] = max(D[y],D[x]+W[y]);
                if (--in[y] == 0)
                {
                    S[++top] = y;
                }
            }
        }
        int ans = 0;
        for (int i = 1; i<=n; i++)
        {
            ans = max(ans,D[i]);
        }
        cout<<ans<<endl;
        return 0;
    }
    

    线性筛素数(真题

    #include<bits/stdc++.h>
    using namespace std;
    int n,m,x;
    bool shai(int a)
    {
        if(a==1) return 0;
        if(a==2 || a==3) return 1;
        if(a%6!=1 && a%6!=5) return 0;
        for(int i=5;i<=sqrt(a);i+=6)
        {
            if(a%i==0 || a%(i+2)==0)
                return 0;
        }
        return 1;
    }
    int main()
    {
        cin>>n>>m;
        for(int i=1;i<=m;i++)
        {
            cin>>x;
            if(shai(x))
                cout<<"Yes"<<endl;
            else
                cout<<"No"<<endl;
            x=0;
        }
        return 0;
    }
    

    最近公共祖先(真题)

    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int n,m,s,tot=0,cnt=0;
    int head[1000100],nxt[1000100],to[1000100];
    int d[500100],f[30][1000100];
    int read()
    {
        int x=0,flag=0;
        char ch=getchar();
        if(ch=='-') flag=1;
        while(ch<'0'||ch>'9')ch=getchar();
        while(ch>='0'&&ch<='9')x*=10,x+=ch-'0',ch=getchar();
        if(flag) return -x;
        return x;
    }
    void addedge(int x,int y)
    {
        cnt++;
        nxt[cnt]=head[x];
        head[x]=cnt;
        to[cnt]=y;
    }
    void dfs(int u,int dep)
    {
        d[u]=dep;
        for(int i=head[u];i!=-1;i=nxt[i])
        {
            int v=to[i];
            if(!d[v]) dfs(v,dep+1),f[0][v]=u;
        }
    }
    int LCA(int x,int y)
    {
        int l=0;
        while((1<<l)<=n) l++;
        l--;
        if(d[x]<d[y]) swap(x, y);
        for(int i=20;i>=0;i--)
            if(d[y]<=d[x]-(1<<i)) x=f[i][x];
        if(x==y) return x;
        for(int i=20;i>=0;i--)
        {
            if(f[i][x]!=f[i][y])
            {
                x=f[i][x];
                y=f[i][y];
            }
        }
        return f[0][x];
    }
    int main()
    {
        memset(head,-1,sizeof(head));
        n=read(),m=read(),s=read();
        for(int i=1;i<n;i++)
        {
            int x,y;
            x=read(),y=read();
            addedge(x,y);
            addedge(y,x);
        }
        f[0][s]=s;
        dfs(s,1);
        for(int i=1;(1<<i)<=n;i++)
            for(int j=1;j<=n;j++)
                f[i][j]=f[i-1][f[i-1][j]];
        for(int i=1;i<=m;i++)
        {
            int l,r;
            l=read(),r=read();
            printf("%d
    ",LCA(l,r));
        }
        return 0;
    }
    

    最长公共子序列(真题

    #include<bits/stdc++.h> 
    using namespace std;
    int a[110000],t[110000],A[110000],B[110000],f[110000];
    bool cmp(int a,int b)
    {
        return a<b;
    }
    int solve(int l,int r,int x)
    {
        int mid=(l+r)/2;;
        if (l==r) return l;
        if (a[mid]>x) 
            return solve(l,mid,x);
        if (a[mid]<=x) 
            return solve(mid+1,r,x);
    }
    int main()
    {
        int n;
        scanf("%d",&n);
        for (int i=1; i<=n; i++) scanf("%d",&A[i]);
        for (int i=1; i<=n; i++) scanf("%d",&B[i]);
        for (int i=1; i<=n; i++) f[A[i]]=i;
        for (int i=1; i<=n; i++) t[i]=f[B[i]];
        memset(a,0,sizeof(a));
        for (int i=1; i<=n; i++)
        {
            if ((i==0)||(t[i]>a[a[0]])) 
                a[++a[0]]=t[i];
            else 
            if (t[i]<a[a[0]]) 
                a[solve(1,a[0],t[i])]=t[i];
        }
        printf("%d
    ",a[0]);
        return 0;
    }
    

    最小生成树(真题

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    using namespace std;
    const int M=200100;
    struct node
    {
        int x;
        int y;
        int w;
    } a[M],t;
    int n,e,dad[M],p=1,ans;
    
    bool cmp(node x,node y)
    {
        if(x.w<y.w)return 1;
        if(x.w==y.w)
            if(x.x>y.x)return 1;
        return 0;
    }
    
    void qsort(int l,int r)
    {
        int i=l,j=r;
        node mid=a[(i+j)/2];
        while(i<=j)
        {
            while(cmp(a[i],mid))i++;
            while(cmp(mid,a[j]))j--;
            if(i<=j)
            {
                t=a[i];
                a[i]=a[j];
                a[j]=t;
                i++;
                j--;
            }
        }
        if(i<r)qsort(i,r);
        if(l<j)qsort(l,j);
    }
    int findx(int x)
    {
        if(x==dad[x])return x;
        dad[x]=findx(dad[x]);
        return dad[x];
    }
    void solve()
    {
        qsort(1,e);
        for(int i=1; i<=n; i++)dad[i]=i;
        for(int i=1; i<=e; i++)
        {
            if(findx(a[i].x)!=findx(a[i].y))
            {
                ans+=a[i].w;
                dad[findx(a[i].x)]=a[i].y;
                p++;
                if(p==n)return ;
            }
        }
    }
    int main()
    {
        cin>>n>>e;
        for(int i=1; i<=e; i++)cin>>a[i].x>>a[i].y>>a[i].w;
        solve();
        cout<<ans;
        return 0;
    }
    

    快读

    long long read()
    {
    	long long x=0,f=1;
    	char ch=getchar();
    	while(ch<'0'||ch>'9')
    	{
    		if(ch=='-')f=-1;
    		ch=getchar();
    	}
    	while(ch>='0'&&ch<='9')
    	{
    		x=x*10+ch-'0';
    		ch=getchar();
    	}
    	return x*f;
    }
    

    严格次小生成树(真题

    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cmath>
    
    using namespace std;
    const int N = 1e5 + 10;
    
    #define gc getchar()
    
    int n, m, now = 1;
    struct Node
    {
        int u, v, w, is_in;
        bool operator <(const Node a)const
        {
            return w < a.w;
        }
    } E[N * 3];
    struct Node_2
    {
        int u, v, w, nxt;
    } G[(N * 3) << 1];
    int fa[N][30], Max[N][30], Tmax[N][30], deep[N], head[N], father[N];
    int Mi[30];
    
    #define LL long long
    LL Answer;
    int Min = 1e9;
    
    inline int read()
    {
        int x = 0;
        char c = gc;
        while(c < '0' || c > '9') c = gc;
        while(c >= '0' || c >= '9') x = x * 10 + c - '0', c = gc;
        return x;
    }
    
    void Add(int u, int v, int w)
    {
        G[now].v = v;
        G[now].w = w;
        G[now].nxt = head[u];
        head[u] = now ++;
    }
    int Getfa(int x)
    {
        return father[x] == x ? x : father[x] = Getfa(father[x]);
    }
    
    void Mst()
    {
        int js(0);
        for(int i = 1; js != n - 1; i ++)
        {
            int u = E[i].u, v = E[i].v, fu = Getfa(u), fv = Getfa(v);
            if(fu != fv)
            {
                father[fu] = fv;
                js ++;
                E[i].is_in = 1;
                Answer += (LL)E[i].w;
                Add(E[i].u, E[i].v, E[i].w);
                Add(E[i].v, E[i].u, E[i].w);
            }
        }
    }
    
    void Dfs(int x, int f_, int dep)
    {
        deep[x] = dep;
        for(int i = 1; ; i ++)
        {
            if(deep[x] - Mi[i] < 0) break;
            fa[x][i] = fa[fa[x][i - 1]][i - 1];
            Max[x][i] = max(Max[x][i - 1], Max[fa[x][i - 1]][i - 1]);
            if(Max[x][i - 1] == Max[fa[x][i - 1]][i - 1])
                Tmax[x][i] = max(Tmax[x][i - 1], Tmax[fa[x][i - 1]][i - 1]);
            else
                Tmax[x][i] = max(min(Max[x][i - 1], Max[fa[x][i - 1]][i - 1]),
                                 max(Tmax[x][i - 1], Tmax[fa[x][i - 1]][i - 1]));
        }
        for(int i = head[x]; ~ i; i = G[i].nxt)
        {
            int v = G[i].v;
            if(v != f_)
            {
                fa[v][0] = x;
                Max[v][0] = G[i].w;
                Tmax[v][0] = -1;
                Dfs(v, x, dep + 1);
            }
        }
    }
    
    int Lca(int x, int y)
    {
        if(deep[x] < deep[y]) swap(x, y);
        int k = deep[x] - deep[y];
        for(int i = 0; i <=  17; i ++)
            if(k >> i & 1) x = fa[x][i];
        if(x == y) return x;
        for(int i = 17; i >= 0; i --)
            if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
        return fa[x][0];
    }
    
    void Work(int s, int t, int w_)
    {
        int m1 = 0, m2 = 0, k = deep[s] - deep[t];
        for(int i = 0; i <= 17; i ++)
        {
            if(k >> i & 1)
            {
                m2 = max(m2, Tmax[s][i]);
                if(Max[s][i] > m1)
                {
                    m2 = max(m2, m1);
                    m1 = Max[s][i];
                }
            }
        }
        if(m1 == w_) Min = min(Min, w_ - m2);
        else Min = min(Min, w_ - m1);
    }
    
    int main()
    {
        n = read();
        m = read();
        for(int i = 1; i <= n; i ++) head[i] = -1, father[i] = i;
        Mi[0] = 1;
        for(int i = 1; i <= 17; i ++) Mi[i] = Mi[i - 1] * 2;
        for(int i = 1; i <= m; i ++)
        {
            E[i].u = read(), E[i].v = read(), E[i].w = read();
        }
        sort(E + 1, E + m + 1);
        Mst();
        Dfs(1, 0, 1);
        for(int i = 1; i <= m; i ++)
        {
            if(!E[i].is_in)
            {
                int u = E[i].u, v = E[i].v;
                int L = Lca(u, v);
                Work(u, L, E[i].w);
                Work(v, L, E[i].w);
            }
        }
        cout << Answer + Min;
        return 0;
    }
    

    最大子段和(真题

    #include <stdlib.h>
    #include<stdio.h>
    int main()
    {
        int count;
        int a[100];
        int b[100];
        int i;
        int max;
        scanf("%d",&count);
        for(i=0; i<count; i++)
        {
            scanf("%d",&a[i]);
        }
        b[0]=a[0];
        max=b[0];
        for(i=1; i<count; i++)
        {
            if(b[i-1]>0)
                b[i]=b[i-1]+a[i];
            else
                b[i]=a[i];
            if(b[i]>max)
                max=b[i];
        }
        printf("%d
    ",max);
        return 0;
    }
    

    快速幂||取余运算(真题

    #include<iostream>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    long long b,p,k,s,t;
    int main()
    {
        cin>>b>>p>>k;
        cout<<b<<"^"<<p<<" mod "<<k<<"=";
        s=b%k;
        t=1;
        for (int i=2;i<=p;i++)
        {
            s=s*b%k;
            if (s==b%k) break;
            t++;
        }
        p=p%t;s=1;
        if (p==0) p=t;
        for (int i=1;i<=p;i++)
        s=s*b%k;
        cout<<s;
        return 0;
    }
    

    快速幂

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cmath>
    using namespace std;
    int ksm(int a,int b,int p)
    {
    	int ans=1;
    	while(b)
    	{
    		if(b&1)				//这一位是不是1 
    			ans=ans*a%p;
    		a=a*a%p;
    		b=b>>1;				//送过来新的一位 
    	}
    	return ans;
    }
    int main()
    {
    	int a,b,p;
    	cin>>a>>b>>p;
    	cout<<ksm(a,b,p);
    	return 0;
    }
    

    卡特兰数

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cmath>
    using namespace std;
    int a[1000010];
    int main()
    {
    	int n;
    	cin>>n;
    	a[0]=1;
    	a[1]=1;
    	for(int i=2;i<=n;i++)
    	{
    		a[i]=a[i-1]*(4*i-2)/(i+1);
    	}
    	cout<<a[n];
    	return 0;
    }
    

    欧拉筛法

    #include <cstring>
    using namespace std;
    int prime[1100000],primesize,phi[11000000];
    bool is[11000000];
    void getlist(int listsize)
    {
    	memset(is,1,sizeof(is));
    	is[1]=false;
    	for(int i=2; i<=listsize; i++)
    	{
    		if(is[i])
    			prime[++primesize]=i;
    		for(int j=1; j<=primesize&&i*prime[j]<=listsize; j++)
    		{
    			is[i*prime[j]]=false;
    			if(i%prime[j]==0)break;
    		}
    	}
    }
    

    Dijkstra

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cmath>
    #include<queue>
    using namespace std;
    struct edge
    {
    	int u,v,x;
    } ed[1000010];
    struct node
    {
    	int x,y;
    	bool operator <(const node & a)const
    	{
    		return y>a.y;
    	}
    };
    priority_queue<node>que;
    int head[1000010];
    int dis[1000010];
    int n,m,s,cnt;
    void add(int x,int y,int z)
    {
    	cnt++;
    	ed[cnt].u=head[x];
    	ed[cnt].v=y;
    	ed[cnt].x=z;
    	head[x]=cnt;
    }
    void dij()
    {
    	for(int i=1; i<=n; i++)
    		dis[i]=2147483647;
    	dis[s]=0;
    	que.push((node)
    	{
    		s,0
    	});
    	while(!que.empty())
    	{
    		node temp=que.top();
    		que.pop();
    		int x=temp.x,y=temp.y;
    		if(y!=dis[x])
    			continue;
    		for(int i=head[x]; i; i=ed[i].u)
    		{
    			if(dis[ed[i].v]>dis[x]+ed[i].x)
    			{
    				dis[ed[i].v]=dis[x]+ed[i].x;
    				que.push((node)
    				{
    					ed[i].v,dis[ed[i].v]
    				});
    			}
    		}
    	}
    }
    int main()
    {
    	cin>>n>>m>>s;
    	for(int i=1; i<=m; i++)
    	{
    		int x,y,z;
    		cin>>x>>y>>z;
    		add(x,y,z);
    	}
    	dij();
    	for(int i=1; i<=n; i++)
    		cout<<dis[i]<<" ";
    	return 0;
    }
    

    单调队列

    #include <cstdio>
    #include <algorithm>
    using namespace std;
    
    const int N = 100000;
    int n, m, head = 1, tail = 0, ans = 2147483647;
    int a[N + 1], f[N + 1], que[N + 1];
    
    int main()
    {
        freopen("beacon.in", "r", stdin);
        freopen("beacon.out", "w", stdout);
    
        scanf("%d%d", &n, &m);
        for (int i = 1;i <= n;i++)
            scanf("%d", &a[i]);
        for (int i = 1;i <= n;i++)
        {
            while (head <= tail && f[i - 1] <= f[que[tail]]) tail--; //当F[i-1]比队尾值更优时把队尾值弹出
            que[++tail] = i - 1; //把F[i-1]插入,这里插入下标而不插入值,便于从队头弹出
            while (head <= tail && que[head] < i - m) head++; //不属于区间维护内的数弹出
            f[i] = f[que[head]] + a[i]; //状态转移
        }
        for (int i = n;i > n - m;i--) //求出答案
            ans = min(ans, f[i]);
        printf("%d
    ", ans);
    
        fclose(stdin);
        fclose(stdout);
        return 0;
    }
    

    EXGCD

    #include<iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<cmath>
    using namespace std;
    int exgcd(int a,int b,int &x,int &y)
    {
        if(b==0)
        {
            x=1,y=0;
            return a;
        }
        int an=exgcd(b,a%b,x,y),temp=x;
        x=y,y=temp-a/b*y;
        return an;
    }
    int main()
    {
        int a,b,x,y;
        cin>>a>>b;
        int ans=exgcd(a,b,x,y);
        while(x<0)
            x+=(b/ans);
        cout<<x;
        return 0;
    }
    

    KMP字符串匹配

    #include<iostream>
    #include<cstring>
    #define MAXN 1000010
    using namespace std;
    int kmp[MAXN];
    int la,lb,j; 
    char a[MAXN],b[MAXN];
    int main()
    {
        cin>>a+1;
        cin>>b+1;
        la=strlen(a+1);
        lb=strlen(b+1);
        for (int i=2;i<=lb;i++)
           {     
           while(j&&b[i]!=b[j+1])
            j=kmp[j];    
           if(b[j+1]==b[i])j++;    
            kmp[i]=j;
           }
        j=0;
        for(int i=1;i<=la;i++)
           {
              while(j>0&&b[j+1]!=a[i])
               j=kmp[j];
              if (b[j+1]==a[i]) 
               j++;
              if (j==lb) {cout<<i-lb+1<<endl;j=kmp[j];}
           }
    
        for (int i=1;i<=lb;i++)
        cout<<kmp[i]<<" ";
        return 0;
    }
    

    字符串哈希

    #include<bits/stdc++.h>
    using namespace std;
    typedef unsigned long long hh;
    hh chang=131;
    hh a[11000];
    char s[11000];
    int n,ans=1;
    hh hashn(char s[])
    {
    	int len=strlen(s);
    	hh ans=0;
    	for(int i=0;i<len;i++)
    		ans=ans*chang+(hh)s[i];
    	return ans & 0x7fffffff;
    }
    int main()
    {
    	cin>>n;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>s;
    		a[i]=hashn(s);
    	}
    	sort(a+1,a+n+1);
    	for(int i=2;i<=n;i++)
    		if(a[i]!=a[i-1])
    		ans++;
    	cout<<ans<<endl;
    	return 0;
    }
    

    缩点(真题

    #include<bits/stdc++.h>
    using namespace std;
    const int N = 110000;
    int dfn[N],low[N],head[N],color[N],sta[N],value[N];
    int f[N],rd[N],dis[N],z[N],x[N],y[N];
    int k=0,n,m,top=0,tim=0,ans=0,ct;
    bool vis[N];
    struct node
    {
        int to,next,from;
    }e[N];
    void add(int u,int v)
    {
        e[++k].to=v;
        e[k].from=u;
        e[k].next=head[u];
        head[u]=k;
    }
    void tarjan(int s)
    {
        dfn[s]=low[s]=++tim;
        sta[++top]=s;
        vis[s]=true;
        for(int i=head[s],v=e[i].to;i;i=e[i].next,v=e[i].to)
        {
            if(!dfn[v])
    			tarjan(v),low[s]=min(low[s],low[v]);
            else if(vis[v])
    			low[s]=min(low[s],dfn[v]);
        }
        if(low[s]==dfn[s])
        {
            ++ct;
            vis[s]=false;
            while(sta[top+1]!=s)
            {
                color[sta[top]]=ct;
                z[ct]+=value[sta[top]];
                vis[sta[top]]=false;
                top--;
            }
        }
    }
    int dfs(int x)
    {
        if(f[x])return f[x];
        f[x]=z[x];
        int maxx=0;
        for(int i=head[x];i;i=e[i].next)
        {
           int to=e[i].to;
           if(!f[to])dfs(to);
           if(maxx>f[to])
    	   maxx=maxx;
    	   else
    	   maxx=f[to];
        }
        f[x]+=maxx;
        return f[x];
    }
    int main()
    {
        cin>>n>>m;
        for(int i=1;i<=n;i++) cin>>value[i];
        for(int i=1;i<=m;i++)
        {
            int u,v;
            cin>>u>>v;
    		x[i]=u,y[i]=v;
            add(u,v);
        }
        for(int i=1;i<=n;i++)
    		if(!dfn[i]) tarjan(i);
        memset(vis,false,sizeof(vis));
        memset(head,0,sizeof(head));
        memset(e,0,sizeof(e));
        k=0;
        for(int i=1;i<=m;i++)
        {
            if(color[x[i]]!=color[y[i]])
            {
                add(color[x[i]],color[y[i]]);
            }
        }
        for(int i=1;i<=ct;++i)
        if(!f[i])
        {
           dfs(i);
           if(ans>f[i])
    	   ans=ans;
    	   else 
    	   ans=f[i];
        }
        cout<<ans;
    }
    

    主席树 (真题

    #include<bits/stdc++.h>
    using namespace std;
    const int MAXN = 210000;
    int n, m, root[MAXN], cut, a[MAXN], s[MAXN], t, x, y, z;
    struct Edge
    {
    	int lc, rc, ans;
    } tree[MAXN * 20];
    void add(int &now, int last, int l, int r, int x)
    {
    	now = ++cut;
    	tree[now].ans = tree[last].ans + 1;
    	tree[now].lc = tree[last].lc, tree[now].rc = tree[last].rc;
    	if(l == r) return ;
    	int mid = (l + r) >> 1;
    	if(x <= mid) add(tree[now].lc, tree[last].lc, l, mid, x);
    	else add(tree[now].rc, tree[last].rc, mid + 1, r, x);
    	return ;
    }
    int query(int L, int R, int l, int r, int x)
    {
    	if(l == r) return l;
    	int p = tree[tree[R].lc].ans - tree[tree[L].lc].ans;
    	int mid = (l + r) >> 1;
    	if(p >= x) return query(tree[L].lc, tree[R].lc, l, mid, x);
    	else return query(tree[L].rc, tree[R].rc, mid + 1, r, x - p);
    }
    int main()
    {
    	cin >> n >> m;
    	for(int i = 1; i <= n; i++) cin >> s[i], a[i] = s[i];
    	sort(s + 1, s + n + 1);
    	for(int i = 1; i <= n; i++)
    	{
    		int p = lower_bound(s + 1, s + n + 1, a[i]) - s;
    		add(root[i], root[i - 1], 1, n, p);
    	}
    	while(m--)
    	{
    		cin >> x >> y >> z;
    		int p = query(root[x - 1], root[y], 1, n, z);
    		printf("%d
    ", s[p]);
    	}
    	return 0;
    }
    

    行来春色三分雨,睡去巫山一片云

  • 相关阅读:
    解析XML技术
    XML名命空间
    XML解析器
    java列表组件鼠标双击事件的实现
    XML(可拓展标记语言)基本概念
    数据包式套接字:基于UDP协议的Socket网络编程
    流式套接字:基于TCP协议的Socket网络编程(案例3)
    剑指 Offer 58
    剑指 Offer 58
    剑指 Offer 57
  • 原文地址:https://www.cnblogs.com/xmex/p/10154800.html
Copyright © 2020-2023  润新知