牛客多校第一场 D Two Graphs（判断图同构）
链接：https://www.nowcoder.com/acm/contest/139/D
来源：牛客网
题目描述
Two undirected simple graphs and where are isomorphic when there exists a bijection on V satisfying if and only if {x, y} ∈ E2.
Given two graphs and , count the number of graphs satisfying the following condition:
* .
* G1 and G are isomorphic.
输入描述:
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains three integers n, m1 and m2 where |E1| = m1 and |E2| = m2.
The i-th of the following m1 lines contains 2 integers ai and bi which denote {ai, bi} ∈ E1.
The i-th of the last m2 lines contains 2 integers ai and bi which denote {ai, bi} ∈ E2.
输出描述:
For each test case, print an integer which denotes the result.
示例1
输入
复制
3 1 2 1 3 1 2 2 3 4 2 3 1 2 1 3 4 1 4 2 4 3
输出
复制
2 3
备注:
* 1 ≤ n ≤ 8 * * 1 ≤ ai, bi ≤ n * The number of test cases does not exceed 50.
题目大意：有两个图
,
，要求出满足如下条件的
的个数：
1、
2、图G与图G1同构
题目思路：由于图中只有8个点，所以我们可以考虑枚举G2的所有的图的情况，再用邻接矩阵来进行判断两个图是否同构，最后还需要减掉G1与G1同构的情况，就是得到最终答案。
#include
#define fi first
#define se second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pb push_back
#define MP make_pair
#define FIN freopen("in.txt","r",stdin)
#define fuck(x) cout<<"["<pii;
const int mod=1e9+7;
int n,m1,m2;
int a[10];
int mp1[10][10],mp2[10][10];
int main(){
while(~scanf("%d%d%d",&n,&m1,&m2)){
memset(mp1,0,sizeof(mp1));
memset(mp2,0,sizeof(mp2));
for(int i=1;i<=n;i++) a[i]=i;
int ans=0,res=0;
for(int i=0;i