AcWing 379. 捉迷藏

題目鏈接：傳送門

DAG的最大獨立集 = 最小路徑覆蓋
最小路徑覆蓋 = 點數 – 最大匹配數
把原圖拆點做做小路徑覆蓋即可
但這個題不用拆點原圖就可以看成二分圖
注意不直接相連的點也算相連
類似於floyed的處理，枚舉中轉點$K$即可
怎麼這麼傳遞閉包

#include <bits/stdc++.h>
#define A 210

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;
}