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 = 30000001, minn;
int a[3001], next[8001], to[8001], head[3001], low[3001], dfn[3001], belong[3001], r[3001];
bool ins[3001], vis[3001], mey[3001];
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] = 1;
    mey[u] = 1;
    for(i = head[u]; i != -1; 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] = 1;
    int i, v;
    for(i = head[u]; i != -1; 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] = 0;
            belong[v] = cnt;
        }while(u != v);
    }
}

int main()
{
    int i, j, k, x, y, u, v;
    scanf("%d %d", &n, &p);
    for(i = 1; i <= p; i++)
    {
        scanf("%d", &x);
        scanf("%d", &a[x]);
    }
    memset(head, -1, sizeof(head));
    scanf("%d", &r1);
    for(i = 1; i <= r1; i++)
    {
        scanf("%d %d", &x, &y);
        add(x, y);
    }
    for(i = 1; i <= n; i++)
     if(!vis[i] && a[i])
      dfs(i);
    for(i = 1; i <= n; i++)
     if(!mey[i])
      ans = min(ans, i);
    if(ans != 30000001)
    {
        printf("NO
%d", ans);
        return 0;
    }
    cnt = 0;
    for(i = 1; i <= n; i++)
     if(!dfn[i])
      tarjan(i);
    for(u = 1; u <= n; u++)
     for(i = head[u]; i != -1; i = next[i])
     {
         v = to[i];
         if(belong[u] != belong[v]) r[belong[v]]++;
     }
    ans = 0;
    for(i = 1; i <= cnt; i++)
     if(r[i] == 0)
     {
         minn = 30000001;
         for(j = 1; j <= n; j++)
          if(belong[j] == i && a[j])
           minn = min(minn, a[j]);
         ans += minn;
     }
    printf("YES
%d", ans);
    return 0;
}