俗话说得好：模板敲得好，NOIP炸不了。
因此为了防止比赛时想到了题解而敲不出来的情况，我们要坚持刷模板。
（争取每次上线的话打1~2个吧）
（排名顺序没有任何意义，想到什么写什么）
SPFA（对应题目Luogu P3371 【模板】单源最短路径）
#include<cstdio> #include<vector> using namespace std; const int N=10005,INF=2147483647; vector <int> a[N],l[N]; int n,m,s,dis[N],i,x,y,z,q[N*20+10]; bool f[N]; inline void read(int &x) { x=0; char ch=getchar(); int flag=1; while (ch<'0'||ch>'9') { if (ch=='-') flag=-1; ch=getchar(); } while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); x*=flag; } int main() { read(n); read(m); read(s); for (i=1;i<=m;++i) { read(x); read(y); read(z); a[x].push_back(y); l[x].push_back(z); } for (i=1;i<=n;++i) dis[i]=INF; dis[s]=0; f[s]=1; q[1]=s; int head=0,tail=1; while (head<tail) { int now=q[++head]; f[now]=0; for (i=0;i<a[now].size();++i) { int k=a[now][i]; if (dis[k]>dis[now]+l[now][i]) { dis[k]=dis[now]+l[now][i]; if (!f[k]) { f[k]=1; q[++tail]=k; } } } } for (i=1;i<=n;++i) printf("%d ",dis[i]); return 0; }
Dijstra（堆优化）（对应题目Luogu P3371 【模板】单源最短路径）
#include<cstdio> #include<vector> #include<queue> using namespace std; const int N=10005,INF=2147483647; vector <int> a[N],l[N]; struct data { int num,s; bool operator < (const data &a) const { return a.s<s; } }; priority_queue <data> small; int n,m,s,dis[N],i,x,y,z,q[N*20+10]; bool f[N]; inline void read(int &x) { x=0; char ch=getchar(); int flag=1; while (ch<'0'||ch>'9') { if (ch=='-') flag=-1; ch=getchar(); } while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); x*=flag; } int main() { read(n); read(m); read(s); for (i=1;i<=m;++i) { read(x); read(y); read(z); a[x].push_back(y); l[x].push_back(z); } for (i=1;i<=n;++i) dis[i]=INF; dis[s]=0; small.push((data){s,0}); while (!small.empty()) { int now=small.top().num; small.pop(); if (f[now]) continue; f[now]=1; for (i=0;i<a[now].size();++i) { int k=a[now][i]; if (dis[k]>dis[now]+l[now][i]) { dis[k]=dis[now]+l[now][i]; small.push((data){k,dis[k]}); } } } for (i=1;i<=n;++i) printf("%d ",dis[i]); return 0; }
线段树（朴素Lazytag）（对应题目 Luogu P3372 【模板】线段树 1）
#include<cstdio> using namespace std; typedef long long LL; const int N=100005; LL seg[N*4+10],add[N*4+10],i,n,m,a[N],x,y,z; inline void read(LL &x) { x=0; char ch=getchar(); int flag=1; while (ch<'0'||ch>'9') { if (ch=='-') flag=-1; ch=getchar(); } while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); x*=flag; } inline void write(LL x) { if (x/10) write(x/10); putchar(x%10+'0'); } inline void up(LL root) { seg[root]=seg[root*2]+seg[root*2+1]; } inline void down(LL root,LL l,LL r) { if (add[root]) { add[root*2]+=add[root]; add[root*2+1]+=add[root]; seg[root*2]+=add[root]*l; seg[root*2+1]+=add[root]*r; add[root]=0; } } inline void build(LL root,LL l,LL r) { if (l==r) seg[root]=a[l]; else { int mid=l+r>>1; build(root*2,l,mid); build(root*2+1,mid+1,r); up(root); } } inline void change(LL root,LL l,LL r,LL ql,LL qr,LL v) { if (l>=ql&&r<=qr) { add[root]+=v; seg[root]+=v*(r-l+1); } else { int mid=l+r>>1; down(root,mid-l+1,r-mid); if (ql<=mid) change(root*2,l,mid,ql,qr,v); if (mid<qr) change(root*2+1,mid+1,r,ql,qr,v); up(root); } } inline LL query(LL root,LL l,LL r,LL ql,LL qr) { if (l>=ql&&r<=qr) return seg[root]; else { int mid=l+r>>1; down(root,mid-l+1,r-mid); LL res=0; if (ql<=mid) res+=query(root*2,l,mid,ql,qr); if (mid<qr) res+=query(root*2+1,mid+1,r,ql,qr); return res; } } int main() { read(n); read(m); for (i=1;i<=n;++i) read(a[i]); build(1,1,n); while (m--) { read(z); if (z==1) { read(x); read(y); read(z); change(1,1,n,x,y,z); } else { read(x); read(y); write(query(1,1,n,x,y)); putchar(' '); } } return 0; }
RMQ（ST表）（对应题目 P3865 【模板】ST表）
#include<bits/stdc++.h> using namespace std; const int N=100005,P=20; int f[N][P],n,m,x,y,i,j; inline void read(int &x) { x=0; char ch=getchar(); int flag=1; while (ch<'0'||ch>'9') { if (ch=='-') flag=-1; ch=getchar(); } while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); x*=flag; } inline void write(int x) { if (x/10) write(x/10); putchar(x%10+'0'); } inline int max(int a,int b) { return a>b?a:b; } int main() { read(n); read(m); for (i=1;i<=n;++i) read(f[i][0]); for (j=1;j<P;++j) for (i=1;i+(1<<j)-1<=n;++i) f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]); while (m--) { read(x); read(y); int mid=(int)log2(y-x+1); write(max(f[x][mid],f[y-(1<<mid)+1][mid])); putchar(' '); } return 0; }
K短路（Astar）（对应题目P2483 【模板】k短路（[SDOI2010]魔法猪学院））
// luogu-judger-enable-o2 #include<cstdio> #include<queue> using namespace std; typedef double DB; const int N=5005; struct data { int num; DB s; bool operator <(const data &a) const { return a.s<s; } }; struct Astar { int num; DB s,rs; bool operator <(const Astar &a) const { return a.s+a.rs<s+rs; } }; priority_queue <data> small; priority_queue <Astar> tree; vector <int> a[N],b[N]; vector <DB> l[N],rl[N]; int n,m,i,x,y,ans; DB z,tot,dis[N],sum; bool vis[N]; inline void read(int &x) { x=0; char ch=getchar(); while (ch<'0'||ch>'9') ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); } int main() { read(n); read(m); scanf("%lf",&tot); for (i=1;i<=m;++i) { read(x); read(y); scanf("%lf",&z); a[x].push_back(y); l[x].push_back(z); b[y].push_back(x); rl[y].push_back(z); } for (i=1;i<=n;++i) dis[i]=1e9; dis[n]=0; small.push((data){n,0}); while (!small.empty()) { int now=small.top().num; small.pop(); if (vis[now]) continue; vis[now]=1; for (i=0;i<b[now].size();++i) { int k=b[now][i]; if (dis[k]>dis[now]+rl[now][i]) { dis[k]=dis[now]+rl[now][i]; small.push((data){k,dis[k]}); } } } tree.push((Astar){1,0,dis[1]}); while (!tree.empty()) { int now=tree.top().num; DB temp=tree.top().s; tree.pop(); if (now==n) { if (sum+temp>tot) { printf("%d",ans); return 0; } else ans++,sum+=temp; } for (i=0;i<a[now].size();++i) tree.push((Astar){a[now][i],temp+l[now][i],dis[a[now][i]]}); } return 0; }
Dinic (对应题目 P3376 【模板】网络最大流）
#include<cstdio> #include<cstring> using namespace std; const int N=10005,M=100005,INF=2e9; struct edge { int to,next,c; }e[M*2]; int q[N],head[N],dep[N],n,m,s,t,i,x,y,z,k=-1; inline void read(int &x) { x=0; char ch=getchar(); while (ch<'0'||ch>'9') ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); } inline void add(int x,int y,int z) { e[++k].to=y; e[k].c=z; e[k].next=head[x]; head[x]=k; } inline int min(int a,int b) { return a<b?a:b; } inline bool BFS(int s,int t) { memset(dep,0,sizeof(dep)); int H=0,T=1; q[1]=s; dep[s]=1; while (H<T) { int now=q[++H]; for (int i=head[now];i!=-1;i=e[i].next) if (!dep[e[i].to]&&e[i].c>0) { dep[e[i].to]=dep[now]+1; q[++T]=e[i].to; } } return dep[t]?1:0; } inline int DFS(int now,int dist) { if (now==t) return dist; int res=0; for (int i=head[now];i!=-1&&dist;i=e[i].next) if (dep[e[i].to]==dep[now]+1&&e[i].c>0) { int dis=DFS(e[i].to,min(e[i].c,dist)); dist-=dis; res+=dis; e[i].c-=dis; e[i^1].c+=dis; } if (!res) dep[now]=0; return res; } inline int Dinic(int s,int t) { int sum=0; while (BFS(s,t)) sum+=DFS(s,INF); return sum; } int main() { memset(e,-1,sizeof(e)); memset(head,-1,sizeof(head)); read(n); read(m); read(s); read(t); for (i=1;i<=m;++i) { read(x); read(y); read(z); add(x,y,z); add(y,x,0); } printf("%d",Dinic(s,t)); return 0; }