蓝题？错误的，这是黄题。

枚举第 $i$ 个左部点，枚举与它匹配的右端点 $j$，判定是否能加入即可。

#include <bits/stdc++.h>
using namespace std;
const int N = 25, M = (1 << 21) + 15, mod = 1e9 + 7;
int n, a[N][N];
int dp[M];

int pop(int S) {
    int res = 0;
    while (S) res += S & 1, S >>= 1;
    return res;
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) scanf("%d", &a[i][j]);
    dp[0] = 1;
    for (int i = 0; i < n; i++) {
        for (int S = 0; S < (1 << n); S++) {
            if (pop(S) != i) continue;
            for (int j = 1; j <= n; j++) {
                if (!a[i + 1][j]) continue;
                if (S & (1 << j - 1)) continue;
                (dp[S | (1 << j - 1)] += dp[S]) %= mod;
            }
        }
    }
    printf("%d\n", dp[(1 << n) - 1]);
    return 0;
}