Two undirected simple graphs G1=⟨V,E1⟩ and G2=⟨V,E2⟩ where V={1,2,⋯,n} are isomorphic when there exists a bijection ϕ on V satisfying {ϕ(x),ϕ(y)}∈E1 if and only if {x,y}∈E2.
Given two graphs G1=⟨V,E1⟩ and G2=⟨V,E2⟩, count the number of graphs G=⟨V,E⟩ satisfying the following condition:
-
E⊆E2.
-
G1 and G are isomorphic.
Input
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.
- 1≤n≤8
- 1≤m1≤m2≤n(n−1)2
- 1≤ai,bi≤n
- The number of test cases does not exceed 50.
Output
For each test case, print an integer which denotes the result.
Example
Input
3 1 2
1 3
1 2
2 3
4 2 3
1 2
1 3
4 1
4 2
4 3
Output
2
3
找图E1的多少子图和E2同构，直接枚举，注意去掉E1自同构的数量
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,m1,m2,x,y,flag,a[10],g1[10][10],g2[10][10];
int fin(int X[10][10],int Y[10][10])//找图X和图Y的同构数
{
for(int i=1;i<=n;i++) a[i]=i;
int ans=0;
do{
flag=1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
if(X[i][j]==1&&Y[a[i]][a[j]]==0)//有一条边连接情况不同，就不是同构
{
flag=0;break;
}
if(!flag) break;
}
ans+=flag;
}while(next_permutation(a+1,a+n+1));
return ans;
}
int main()
{
while(scanf("%d%d%d",&n,&m1,&m2)==3)
{
memset(g1,0,sizeof(g1));
memset(g2,0,sizeof(g2));
for(int i=0;i<m1;i++)
{
scanf("%d%d",&x,&y);
g1[x][y]=g1[y][x]=1;
}
for(int i=0;i<m2;i++)
{
scanf("%d%d",&x,&y);
g2[x][y]=g2[y][x]=1;
}
int ans=fin(g1,g2),sum=fin(g1,g1);
printf("%d\n",ans/sum);
}
return 0;
}