题意

n个点，m1条边的图E1，n个点，m2条边的图E2。求图E2有多少子图跟图E1同构。

题解

由于题目数据范围不大考虑暴力枚举$n!$范围的所有结果。
答案除以自同构数。

AC代码

#include <iostream>
#include <cstring>
#include <algorithm>
#include <numeric>

using namespace std;
int n, m1, m2, u[100], v[100], g1[10][10], g2[10][10], a[10];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    while (cin >> n >> m1 >> m2) {
        memset(g1, 0, sizeof(g1));
        memset(g2, 0, sizeof(g2));
        for (int i = 1; i <= m1; i++) {
            cin >> u[i] >> v[i];
            g1[u[i]][v[i]] = g1[v[i]][u[i]] = 1;
        }
        for (int i = 1; i <= m2; i++) {
            int x, y;
            cin >> x >> y;
            g2[x][y] = g2[y][x] = 1;
        }
        iota(a + 1,a + 1 + n,1);
        int ans = 0, num = 0;
        do {
            int flag1 = 1, flag2 = 1;
            for (int i = 1; i <= m1; i++) {
                int x = a[u[i]], y = a[v[i]];
                if (!g1[x][y])flag1 = 0;
                if (!g2[x][y])flag2 = 0;
            }
            ans += flag2;
            num += flag1;
        } while (next_permutation(a + 1, a + n + 1));
        cout << ans / num << "\n";
    }
}