Unbearable Controversy of Being CodeForces - 489D
2017-12-09 11:05
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这题真是看你敢不敢想 纯暴力 枚举再枚举
数据量再小一点的话就用反向建边 可达矩阵相乘也行
这题真是看你敢不敢想 纯暴力 枚举再枚举
数据量再小一点的话就用反向建边 可达矩阵相乘也行
#include <bits/stdc++.h> using namespace std; struct node { int v; int next; }; node edge[30010]; int first[3010],book[3010]; int n,m,num; void addedge(int u,int v) { edge[num].v=v; edge[num].next=first[u]; first[u]=num++; return; } int main() { int i,j,u,v,w,ans; while(scanf("%d%d",&n,&m)!=EOF) { memset(first,-1,sizeof(first)); num=0; for(i=1;i<=m;i++) { scanf("%d%d",&u,&v); addedge(u,v); } ans=0; for(u=1;u<=n;u++) { memset(book,0,sizeof(book)); for(i=first[u];i!=-1;i=edge[i].next) { v=edge[i].v; for(j=first[v];j!=-1;j=edge[j].next) { w=edge[j].v; if(w!=u) { book[w]++; } } } for(i=1;i<=n;i++) { ans+=(book[i]*(book[i]-1))/2; } } printf("%d\n",ans); } return 0; }
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