POJ2367（拓扑排序裸题

 1 #include<iostream>
 2 #include<vector>
 3 #include<queue>
 4 using namespace std;
 5 typedef long long ll;
 6 const int N = 150;
 7 int in[N],n,r;
 8 vector<int>ans;
 9 vector<int>edge[N];
10 priority_queue<int,vector<int>,greater<int> >priq;
11 int main(){
12     ios::sync_with_stdio(0);
13     cin>>n;
14     for(int i = 1;i <= n;++i){
15         while(cin>>r){
16             if(r==0)break;
17             edge[i].push_back(r);in[r]++;
18         }
19     }
20     for(int i = 1;i <= n;++i)if(in[i]==0)priq.push(i);
21     while(!priq.empty()){
22         
23         int p = priq.top();priq.pop();
24         ans.push_back(p);
25         for(int j = 0;j < edge[p].size();++j){
26             int y = edge[p][j];in[y]--;
27             if(!in[y])priq.push(y);
28         }
29     }
30     for(int i = 0;i < n;++i){
31         if(i!=n-1)cout<<ans[i]<<" ";
32         else cout<<ans[i]<<endl;
33     }
34     return 0;
35 }