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.#include<bits/stdc++.h>
using namespace std;
#define ll long longint n,m1,m2;
struct edge{ int a , b ; };
using graph=vector<edge> ;
/*
题目大意：给定两张图，a,b,
求在图b中和a 同构的子图个数。关于同构的判断处理方法是：
把下标进行排列映射，带入判断。
*/
graph read(int m)
{graph ans;for(int i=0,a,b;i<m;i++){scanf("%d%d",&a,&b); a--,b--;ans.push_back( edge{a,b} );}return ans;
}int Count(int n,graph &a,graph &b)
{vector< vector<bool> > target( n,vector<bool>(n,false) );for(auto && e : b ) target[e.a][e.b]=target[e.b][e.a]=true;vector<int> phi(n);iota(phi.begin(),phi.end(),0);///利用了函数iota方法，批量递增int ans=0;do{bool k=true;for(auto &&e :a) k&=target[phi[e.a]][phi[e.b]];ans+=k;}while(next_permutation( phi.begin() , phi.end() ) );return ans;
}int main()
{while(~scanf("%d%d%d",&n,&m1,&m2)){graph A=read(m1);graph B=read(m2);int isab=Count(n,A,B);int isaa=Count(n,A,A);///记得要去除自同构的个数printf("%d\n",isab/isaa);}return 0;
}