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题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
#include <bits/stdc++.h> using namespace std; const int N=3e4+5; typedef long long ll; const int mod=1e9+7; int n,m,ind[N],u,v,ans[N]; priority_queue<int>qu; vector<int>ve[N]; int main() { int t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { ve[i].clear(); ind[i]=0; } for(int i=1;i<=m;i++) { scanf("%d%d",&u,&v); ve[v].push_back(u); ind[u]++; } int cnt=n; for(int i=1;i<=n;i++) { if(!ind[i])qu.push(i); } while(!qu.empty()) { int fr=qu.top(); ans[cnt--]=fr; qu.pop(); int len=ve[fr].size(); for(int i=0;i<len;i++) { int y=ve[fr][i]; ind[y]--; if(!ind[y])qu.push(y); } } for(int i=1;i<n;i++) { printf("%d ",ans[i]); } printf("%d\n",ans[n]); } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5402843.html