Two graph

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

Two Graphs

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld

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.

#include<bits/stdc++.h>
using namespace std;

int n,m1,m2,a,b;
int vis[10][10],pre[10],w[10],ans;
map<long long,int>mm;

struct AA
{
    int x,y;
}pos[50];

int dfs(int rt)///枚举所有的点
{
    if(rt==n+1)
    {
        long long pp=0;
        for(int i=1;i<=m1;i++)
        {
            if(vis[pre[pos[i].x]][pre[pos[i].y]]==0)
                return 0;
            pp|=1<<vis[pre[pos[i].x]][pre[pos[i].y]];
        }
        if(mm[pp])
            return 0;//判断选择的E2的边是否重复
        ans++;
        mm[pp]=1;
        return 0;
    }
    for(int i=1;i<=n;i++)
    {
        if(w[i])
            continue;
        pre[rt]=i;//E1中的rt点对应E2中的i点，映射
        w[i]=1;
        dfs(rt+1);
        w[i]=0;
    }
    return 0;
}
int main()
{
    while(~scanf("%d%d%d",&n,&m1,&m2))
    {
        mm.clear();
        memset(vis,0,sizeof(vis));
        memset(w,0,sizeof(w));
        ans=0;
        int x,y;
        for(int i=1;i<=m1;i++)
        {
            scanf("%d%d",&x,&y);
            pos[i].x=x;///E1中边的端点
            pos[i].y=y;
        }
        for(int i=1;i<=m2;i++)
        {
            scanf("%d%d",&x,&y);
            vis[x][y]=vis[y][x]=i;///E2中的边的编号
        }
        dfs(1);
        printf("%d\n",ans);
    }
    return 0;
}