【提高组】倍增

P1967 货车运输

*Kruskal重构树模板题。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<vector>
#include<stack>
#include<cmath>
#include<bits/stdc++.h>
#define ri register int
#define ll long long
#define For(i,l,r) for(ri i=l;i<=r;i++)
#define Dfor(i,r,l) for(ri i=r;i>=l;i--)
using namespace std;
const int M=200010;
int n,m,cnt,q,fa[M],tot,head[M],val[M],size[M],top[M],son[M],dep[M],f[M];
bool vis[M];
struct node{
    int u,v,dis;
    bool operator < (const node&x) const{
        return dis<x.dis;
    }
}e[M];
inline bool cmp(node x,node y){return x.dis>y.dis;}
struct Node{
    int nxt,to;
}ee[M];
inline ll read(){
    ll f=1,sum=0;
    char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
    return f*sum;
}
inline int getfa(int x){
    if(x==fa[x]) return x;
    return fa[x]=getfa(fa[x]);
}
inline void add(int u,int v){
    ee[++tot].nxt=head[u];
    head[u]=tot;
    ee[tot].to=v;
}
inline void dfs1(int u,int pa){
    size[u]=1;vis[u]=1;
    for(ri i=head[u];i;i=ee[i].nxt){
        int v=ee[i].to;
        if(v==pa)continue;
        dep[v]=dep[u]+1;f[v]=u;
        dfs1(v,u);
        size[u]+=size[v];
        if(size[v]>size[son[u]]) son[u]=v;
    }
}
inline void dfs2(int u,int tp){
    top[u]=tp;
    if(son[u]) dfs2(son[u],tp);
    for(ri i=head[u];i;i=ee[i].nxt){
        int v=ee[i].to;
        if(v==son[u]||v==f[u]) continue;
        dfs2(v,v);
    }
}
inline int lca(int u,int v){
    while(top[u]!=top[v]){
        if(dep[top[u]]>dep[top[v]]) u=f[top[u]];
        else v=f[top[v]];
    }
    return dep[u]<dep[v]?u:v;
}
inline void kruskal(){
    For(i,1,n) fa[i]=i;
    sort(e+1,e+m+1,cmp);
    For(i,1,m){
        int fu=getfa(e[i].u),fv=getfa(e[i].v);
        if(fu!=fv){
            val[++cnt]=e[i].dis;
            fa[cnt]=fa[fu]=fa[fv]=cnt;
            add(fu,cnt),add(cnt,fu);
            add(fv,cnt),add(cnt,fv);
        }
    }
    For(i,1,cnt){
        if(!vis[i]){
            int ff=getfa(i);
            dfs1(ff,0),dfs2(ff,ff);
        }
    }
}
int main(){
    n=read(),m=read(),cnt=n;
    For(i,1,m){
        e[i].u=read(),e[i].v=read(),e[i].dis=read();
    }
    kruskal();
    q=read();
    while(q--){
        int u=read(),v=read();
        if(getfa(u)!=getfa(v)) printf("-1
");
        else printf("%d
",val[lca(u,v)]);
    }
    return 0;
}

*倍增

P1081 开车旅行

*俩神经病换着还要凭喜好开车的故事...怎么跟小学的憨憨放水进水的奥数题一样...

*70'的暴力我卡在了初始化上，我是想预处理出各个城市的距离再sort，题解是

 枚举其实从1到n-1和n-1到1都没有区别。提前处理出四个需要用到的量，其他都不用管这个idea很好。看下我被卡的代码实现，主要是没想到只处理出4个量，按我原来的sort实现不了。

For(i,1,n-1){
        ll minn1=i+1,minn2=0;
        dis1[i]=abs(h[i]-h[i+1]);
        For(j,i+2,n){
            if(dis1[i]>abs(h[i]-h[j])||dis1[i]==abs(h[i]-h[j])&&h[j]<h[minn1]){
                dis2[i]=dis1[i],dis1[i]=abs(h[i]-h[j]);
                minn2=minn1,minn1=j;
            }
            else if(!dis2[i]||dis2[i]>abs(h[i]-h[j])||dis2[i]==abs(h[i]-h[j])&&h[j]<h[minn2]){
                dis2[i]=abs(h[i]-h[j]);
                minn2=j;
            }
        }
        c1[i]=minn1,c2[i]=minn2;
    }

换车那里记录一个turn=1/0，表示a/b开车，统计一下判断是否符合条件即可。

*70'代码：

#include<bits/stdc++.h>
#define ri register int
#define ll long long
#define For(i,l,r) for(ri i=l;i<=r;i++)
#define Dfor(i,r,l) for(ri i=r;i>=l;i--)
using namespace std;
const int M=100010;
ll n,m,xo,ans,c1[M],c2[M],dis1[M],dis2[M],h[M];
double minv=1e9;
inline ll read(){
    ll f=1,sum=0;
    char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
    return f*sum;
}
inline void write(ll x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
int main(){
    n=read();
    For(i,1,n) h[i]=read();
    For(i,1,n-1){
        ll minn1=i+1,minn2=0;
        dis1[i]=abs(h[i]-h[i+1]);
        For(j,i+2,n){
            if(dis1[i]>abs(h[i]-h[j])||dis1[i]==abs(h[i]-h[j])&&h[j]<h[minn1]){
                dis2[i]=dis1[i],dis1[i]=abs(h[i]-h[j]);
                minn2=minn1,minn1=j;
            }
            else if(!dis2[i]||dis2[i]>abs(h[i]-h[j])||dis2[i]==abs(h[i]-h[j])&&h[j]<h[minn2]){
                dis2[i]=abs(h[i]-h[j]);
                minn2=j;
            }
        }
        c1[i]=minn1,c2[i]=minn2;
    }
    xo=read(),ans=0;
    For(i,1,n){
        ll a=0,b=0,loc=i,turn=0;
        while(1){
            if(turn){
                if(a+b+dis1[loc]>xo||!c1[loc]) break;
                b+=dis1[loc];
                loc=c1[loc];
            }
            else{
                if(a+b+dis2[loc]>xo||!c1[loc]) break;
                a+=dis2[loc];
                loc=c2[loc];
            }
            turn^=1;
        }
        if(!ans||1.0*a/b-minv<-0.00000001||fabs((1.0*a/b-minv))<=0.00000001&&h[ans]<h[i]){
            minv=1.0*a/b;
            ans=i;
        }
    }
    write(ans);printf("
");
    m=read();
    while(m--){
        ll s=read(),x=read(),a=0,b=0,turn=0;
        while(1){
            if(turn){
                if(a+b+dis1[s]>x||!c1[s]) break;
                b+=dis1[s];
                s=c1[s];
            }
            else{
                if(a+b+dis2[s]>x||!c1[s]) break;
                a+=dis2[s];
                s=c2[s];
            }
            turn^=1;
        }
        write(a);printf(" ");write(b);printf("
");
    }
    return 0;
}

 *AC代码就是用双向链表初始化+倍增优化，就是多两个优化啦。不多赘述，不会实现的话看->https://www.luogu.org/blog/Zechariah/solution-p1081。

*AC代码：

#include<bits/stdc++.h>
#define ri register int
#define ll long long
#define For(i,l,r) for(ri i=l;i<=r;i++)
#define Dfor(i,r,l) for(ri i=r;i>=l;i--)
using namespace std;
const int M=100010;
ll c1[M],c2[M],c3[M][21],n,m,pos[M],dis1[M],dis2[M],dis3[M][21],dis4[M][21],dis5[M][21];
double minv=1e9;
struct node {
    ll h;
    int bh,last,next;
}q[M];
inline ll read(){
    ll f=1,sum=0;
    char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
    return f*sum;
}
inline void write(ll x){
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
inline bool cmp(node s,node t){return s.h<t.h;}
inline void updata(int i,int loc,int x){
    if(x>=1&&x<=n){
        if(!dis1[i]||dis1[i]>abs(q[loc].h-q[x].h)||(dis1[i]==abs(q[loc].h-q[x].h)&&q[x].h<q[pos[c1[i]]].h)){
            dis2[i]=dis1[i];
            dis1[i]=abs(q[loc].h-q[x].h);
            c2[i]=c1[i];
            c1[i]=q[x].bh;
        }
        else if(!dis2[i]||dis2[i]>abs(q[loc].h-q[x].h)||(dis2[i]==abs(q[loc].h-q[x].h)&&q[x].h<q[pos[c2[i]]].h)){
            dis2[i]=abs(q[loc].h-q[x].h);
            c2[i]=q[x].bh;
        }
    }
}

int main(){
    n=read();
    For(i,1,n) q[i].h=read(),q[i].bh=i;
    sort(q+1,q+n+1,cmp);
    For(i,1,n){ 
        pos[q[i].bh]=i;
        if(i!=1)q[i].last=i-1;
        if (i!=n)q[i].next=i+1;
    }
    For(i,1,n){
        int loc = pos[i];
        updata(i,loc,q[q[loc].last].last);
        updata(i,loc,q[loc].last);
        updata(i,loc,q[loc].next);
        updata(i,loc,q[q[loc].next].next);
        if(q[loc].last)q[q[loc].last].next=q[loc].next;
        if(q[loc].next)q[q[loc].next].last=q[loc].last;
        q[loc].last=q[loc].next=0;
    }
    For(i,1,n){
        dis4[i][0]=dis2[i];
        dis5[i][0]=dis1[c2[i]];
        dis3[i][0]=dis2[i]+dis1[c2[i]];
        c3[i][0]=c1[c2[i]];
    }
    For(j,1,20)
        For(i,1,n){
            c3[i][j]=c3[c3[i][j-1]][j - 1];
            if(c3[i][j]){
                dis3[i][j]=dis3[i][j-1]+dis3[c3[i][j-1]][j-1];
                dis4[i][j]=dis4[i][j-1]+dis4[c3[i][j-1]][j-1];
                dis5[i][j]=dis5[i][j-1]+dis5[c3[i][j-1]][j-1];
            }
        }
    ll xx=read(),ans=0;
    For(i,1,n){
        ll a=0,b=0,loc=i,x0=xx;
        Dfor(j,20,0){
            if(dis3[loc][j]&&x0>=dis3[loc][j]){
                x0-=dis3[loc][j];
                a+=dis4[loc][j];
                b+=dis5[loc][j];
                loc=c3[loc][j];
            }
        }
        if(dis2[loc]<=x0)a+=dis2[loc];
        if(a<=0)continue;
        if(!ans||1.0*a/b-minv<-0.00000001||(fabs(1.0*a/b-minv)<=0.00000001&&q[pos[ans]].h<q[pos[i]].h)){
            minv=1.0*a/b;
            ans=i;
        }
    }
    write(ans);printf("
");
    m=read();
    while(m--){
        ll s=read(),x=read(),a=0,b=0;
        Dfor(j,20,0){
            if(dis3[s][j]&&x>=dis3[s][j])
            {
                x-=dis3[s][j];
                a+=dis4[s][j];
                b+=dis5[s][j];
                s=c3[s][j];
            }
        }
        if(dis2[s]<=x)a+=dis2[s];
        write(a);printf(" ");write(b);printf("
");
    }
    return 0;
}

P1613 跑路

*题面简直就是清清楚楚得把倍增拍你脸上了...2^k都出来了（然而接近10点搞了一天的我却瞎了眼

*直接放代码吧

#include<bits/stdc++.h>
#define ri register int
#define ll long long
#define For(i,l,r) for(ri i=l;i<=r;i++)
#define Dfor(i,r,l) for(ri i=r;i>=l;i--)
using namespace std;
int dis[60][60],n,m;
bool g[60][60][65];
inline ll read(){
    ll f=1,sum=0;
    char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();}
    return f*sum;
}
int main(){
    n=read();m=read();
    memset(dis,10,sizeof(dis));
    For(i,1,m){
        int u=read(),v=read();
        dis[u][v]=1;
        g[u][v][0]=1;
    }
    For(k,1,64){
        For(i,1,n){
            For(t,1,n){
                For(j,1,n){
                    if(g[i][t][k-1]&&g[t][j][k-1]) dis[i][j]=1,g[i][j][k]=1;
                }
            }
        }
    }
    For(k,1,n){
        For(i,1,n){
            For(j,1,n){
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
            }
        }
    }
    printf("%d
",dis[1][n]);
    return 0;
}