间谍网络（tarjan缩点）

洛谷传送门

看着这道题给人感觉就是tarjan求SCC，然而还得判断是否能控制全部间谍，这就得先从可以贿赂的点dfs一遍。

如果没有全部被标记了，就输出NO，再从没被标记的点里找最小的标号。

如果全被标记，就输出YES，再从入度为0的缩点里找最小的价格，加到ans中，最后输出ans。

——代码

 #include <cstdio>
 #include <stack>
 #include <cstring>

 using namespace std;

 int n, p, r1, cnt, idx, ans = , minn;
 int a[], next[], to[], head[], low[], dfn[], belong[], r[];
 bool ins[], vis[], mey[];
 stack <int> s;

 inline void add(int x, int y)
 {
     to[cnt] = y;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 void dfs(int u)
 {
     int i, v;
     vis[u] = ;
     mey[u] = ;
     for(i = head[u]; i != -; i = next[i])
     {
         v = to[i];
         if(!vis[v]) dfs(v);
     }
 }

 void tarjan(int u)
 {
     low[u] = dfn[u] = ++idx;
     s.push(u);
     ins[u] = ;
     int i, v;
     for(i = head[u]; i != -; i = next[i])
     {
         v = to[i];
         if(!dfn[v])
         {
             tarjan(v);
             low[u] = min(low[u], low[v]);
         }
         else if(ins[v]) low[u] = min(low[u], dfn[v]);
     }
     if(low[u] == dfn[u])
     {
         cnt++;
         do
         {
             v = s.top();
             s.pop();
             ins[v] = ;
             belong[v] = cnt;
         }while(u != v);
     }
 }

 int main()
 {
     int i, j, k, x, y, u, v;
     scanf("%d %d", &n, &p);
     for(i = ; i <= p; i++)
     {
         scanf("%d", &x);
         scanf("%d", &a[x]);
     }
     memset(head, -, sizeof(head));
     scanf("%d", &r1);
     for(i = ; i <= r1; i++)
     {
         scanf("%d %d", &x, &y);
         add(x, y);
     }
     for(i = ; i <= n; i++)
      if(!vis[i] && a[i])
       dfs(i);
     for(i = ; i <= n; i++)
      if(!mey[i])
       ans = min(ans, i);
     if(ans != )
     {
         printf("NO\n%d", ans);
         return ;
     }
     cnt = ;
     for(i = ; i <= n; i++)
      if(!dfn[i])
       tarjan(i);
     for(u = ; u <= n; u++)
      for(i = head[u]; i != -; i = next[i])
      {
          v = to[i];
          if(belong[u] != belong[v]) r[belong[v]]++;
      }
     ans = ;
     for(i = ; i <= cnt; i++)
      if(r[i] == )
      {
          minn = ;
          for(j = ; j <= n; j++)
           if(belong[j] == i && a[j])
            minn = min(minn, a[j]);
          ans += minn;
      }
     printf("YES\n%d", ans);
     return ;
 }