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.
题解:判断同构:判断两个图的邻接矩阵是否一样就行了,枚举G2的所有可能可能情况,用全排列枚举,状压存下来,再放在一个set去重
代码:
#include<bits/stdc++.h>
using namespace std;
int n,m2,m1;
#define MAX 10
#define pii pair<int,int>
typedef long long ll;
int G1[MAX][MAX],G2[MAX][MAX];
map<pii,int>mp;
set<ll>s;
int main()
{
while(cin>>n>>m1>>m2){
mp.clear();
s.clear();
memset(G1,0,sizeof(G1));
memset(G2,0,sizeof(G2));
for(int i=0;i<m1;i++)
{
int u,v;
cin>>u>>v;
G1[u][v]=G1[v][u]=1;
}
for(int i=0;i<m2;i++)
{
int u,v;
cin>>u>>v;
G2[u][v]=G2[v][u]=1;
if(u<v)swap(u,v);
mp[make_pair(u,v)]=i;
}
ll ans[MAX];
for(int i=1;i<=n;i++)
ans[i]=i;
do{
ll state=0;
bool ok=true;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(G1[i][j]&&!G2[ans[i]][ans[j]])
{
ok=false;
break;
}
else
{
if(G1[i][j])
{
int u=ans[i];
int v=ans[j];
if(u<v)swap(u,v);
state|=(1LL<<mp[make_pair(u,v)]);
}
}
}
if(!ok)break;
}
if(!ok)continue;
s.insert(state);//cout<<s.size()<<"00000"<<endl;
}while(next_permutation(ans+1,ans+1+n));
cout<<s.size()<<endl;
}
}