AcWing 379. 捉迷藏

题目链接：​​传送门​​

DAG的最大独立集 = 最小路径覆盖
最小路径覆盖 = 点数 – 最大匹配数
把原图拆点做做小路径覆盖即可
但这个题不用拆点原图就可以看成二分图
注意不直接相连的点也算相连
类似于floyed的处理，枚举中转点即可
怎么这么传递闭包

#include <bits/stdc++.h>
#define

using namespace std;
int n, m, a, b, g[A], ans; bool ap[A][A], vis[A];
bool check(int x) {
  for (int i = 1; i <= n; i++)
    if (ap[x][i] and !vis[i]) {
      vis[i] = 1;
      if (!g[i] or check(g[i])) {
        g[i] = x;
        return true;
      }
    }
  return false;
}

int main(int argc, char const *argv[]) {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) scanf("%d%d", &a, &b), ap[a][b] = 1;
  for (int k = 1; k <= n; k++)
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (i != j and j != k)
        ap[i][j] |= ap[i][k] && ap[k][j];
  for (int i = 1; i <= n; i++) {
    memset(vis, 0, sizeof vis);
    if (check(i)) ans++;
  }
  cout << n - ans << endl;
  return 0;
}