先找出强连通分量缩点，然后就是最小路径覆盖。
构造一个二分图，把每个点\(i\)拆成两个点\(X_i,Y_i\)。
对于原图中的边\(u \to v\)，在二分图添加一条边\(X_u \to Y_v\)。

• 最小路径覆盖 = 顶点个数 - 最大匹配数

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std;const int maxn = 5000 + 10;
const int maxm = 100000 + 10;struct Edge
{int v, nxt;Edge() {}Edge(int v, int nxt): v(v), nxt(nxt) {}
};int ecnt, head[maxn];
Edge edges[maxm];void AddEdge(int u, int v) {edges[ecnt] = Edge(v, head[u]);head[u] = ecnt++;
}int n, m;stack<int> S;
int dfs_clock, pre[maxn], low[maxn];
int scc_cnt, sccno[maxn];void dfs(int u) {pre[u] = low[u] = ++dfs_clock;S.push(u);for(int i = head[u]; ~i; i = edges[i].nxt) {int v = edges[i].v;if(!pre[v]) {dfs(v);low[u] = min(low[u], low[v]);} else if(!sccno[v]) low[u] = min(low[u], pre[v]);}if(low[u] == pre[u]) {scc_cnt++;for(;;) {int x = S.top(); S.pop();sccno[x] = scc_cnt;if(x == u) break;}}
}void find_scc() {dfs_clock = scc_cnt = 0;memset(pre, 0, sizeof(pre));memset(sccno, 0, sizeof(sccno));for(int i = 1; i <= n; i++) if(!pre[i])dfs(i);
}int ecnt2, head2[maxn];
Edge edges2[maxm];
int left[maxn];
bool vis[maxn];void AddEdge2(int u, int v) {edges2[ecnt2] = Edge(v, head2[u]);head2[u] = ecnt2++;
}bool find(int u) {for(int i = head2[u]; ~i; i = edges2[i].nxt) {int v = edges2[i].v;if(vis[v]) continue;vis[v] = true;if(!left[v] || find(left[v])) {left[v] = u;return true;}}return false;
}int main()
{int T; scanf("%d", &T);while(T--) {scanf("%d%d", &n, &m);ecnt = 0;memset(head, -1, sizeof(head));while(m--) {int u, v; scanf("%d%d", &u, &v);AddEdge(u, v);}find_scc();ecnt2 = 0;memset(head2, -1, sizeof(head2));for(int u = 1; u <= n; u++) {for(int i = head[u]; ~i; i = edges[i].nxt) {int v = edges[i].v;if(sccno[u] == sccno[v]) continue;AddEdge2(sccno[u], sccno[v]);}}int match = 0;memset(left, 0, sizeof(left));for(int i = 1; i <= scc_cnt; i++) {memset(vis, 0, sizeof(vis));if(find(i)) match++;}printf("%d\n", scc_cnt - match);}return 0;
}