我们首先考虑该题没有环应该怎么做，因为没有环所以是一个DAG，因此直接加上入度为0的罪犯，而有环则可以缩点，之后就成为了DAG，然后用一方法做就好了。
\(Code\)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int n, p, r, top, point, cnt, color, ans, belong[100100], stack[100100], lin[100100], low[100010], dfn[100100], data[100100], vis[100100], minn[100010], in[100100];
// 共有几个间谍， 被收买的间谍的值，
struct edge {
int from, to, nex;
}e[100100];
inline void add(int f, int t)
{
e[++cnt].from = f;
e[cnt].to = t;
e[cnt].nex = lin[f];
lin[f] = cnt;
}
void tarjan(int u)
{
low[u] = dfn[u] = ++point;
stack[++top] = u;
vis[u] = 1;
for(int i = lin[u]; i; i = e[i].nex)
{
int v = e[i].to;
if(!dfn[v]) tarjan(v), low[u] = min(low[u],low[v]);
else if(vis[v]) low[u] = min(low[u],dfn[v]);
}
if(low[u] == dfn[u])
{
color++;
int v = 0;
while(v != u)
{
v = stack[top--];
vis[v] = 0;
belong[v] = color;
minn[color] = min(minn[color], data[v]);
}
}
}
int main()
{
scanf("%d%d", &n, &p);
for (int i = 1; i <= n; i++)
data[i] = 2147483647, minn[i] = 2147483647;
for (int i = 1; i <= p; i++)
{
int a, b;
scanf("%d%d", &a, &b);
data[a] = b;
}
scanf("%d", &r);
for (int i = 1; i <= r; i++)
{
int a, b;
scanf("%d%d", &a, &b);
add(a, b);
}
for (int i = 1; i <= n; i++)
if (!dfn[i] && data[i] != 2147483647) tarjan(i);
for (int i = 1; i <= n; i++)
if (!dfn[i])
printf("NO\n%d", i), exit(0);
printf("YES\n");
for (int cur = 1; cur <= n; cur++)
for (int i = lin[cur]; i; i = e[i].nex)
{
int to = e[i].to;
if (belong[cur] != belong[to])
in[belong[to]]++;
}
for (int i = 1; i <= color; i++)
if (!in[i])
ans += minn[i];
printf("%d", ans);
}