牛客网暑期ACM多校训练营（第一场）D Two Graphs

链接：https://www.nowcoder.com/acm/contest/139/D
来源：牛客网

题目描述

Two undirected simple graphs $G_1 = \langle V, E_1 \rangle$ and $G_2 = \langle V, E_2 \rangle$ where $V = \{1, 2, \dots, n\}$ are isomorphic when there exists a bijection $\phi$ on V satisfying $\{\phi(x), \phi(y)\} \in E_1$ if and only if {x, y} ∈ E2.
Given two graphs $G_1 = \langle V, E_1 \rangle$ and $G_2 = \langle V, E_2 \rangle$ , count the number of graphs $G = \langle V, E \rangle$ satisfying the following condition:
* $E \subseteq E_2$ .
* 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

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：给两个图，判断图2有多少个子图与图1同构

用全排列枚举所有的情况，然后还要算一遍自同构，也就是说如两个图都是 1-2,1-3，看上去是有两种映射方式，但实际上我们是算成一种

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=10;
int mapf[maxn][maxn];
int maps[maxn][maxn];
int num[maxn];
int main()
{
    int n,m1,m2;
    while(~scanf("%d%d%d",&n,&m1,&m2))
    {
    	memset(mapf,0,sizeof(mapf));
    	memset(maps,0,sizeof(maps));
        for(int i=0;i<m1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            mapf[u][v]=mapf[v][u]=1;
        }
        for(int i=0;i<m2;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            maps[u][v]=maps[v][u]=1;
        }
        for(int i=1;i<=n;i++)
        {
            num[i]=i;
        }
        int ans=0;
        do
        {
        	int flag=1;
        	for(int i=1;i<=n;i++)
			{
				if(!flag) break;
				for(int j=1;j<=n;j++)
				{
					if(mapf[i][j]==1&&maps[num[i]][num[j]]==0)
					{
						flag=0;
						break;
					}
				}
			}
			ans+=flag;
        }while(next_permutation(num+1,num+n+1));
        for(int i=1;i<=n;i++)
        {
            num[i]=i;
        }
        int res=0;
        do
        {
        	int flag=1;
        	for(int i=1;i<=n;i++)
			{
				if(!flag) break;
				for(int j=1;j<=n;j++)
				{
					if(mapf[i][j]==1&&mapf[num[i]][num[j]]==0)
					{
						flag=0;
						break;
					}
				}
			}
			res+=flag;
        }while(next_permutation(num+1,num+n+1));
        cout<<ans/res<<endl;
    }
    return 0;
}