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
3 1 2
1 3
1 2
2 3
4 2 3
1 2
1 3
4 1
4 2
4 3
输出
2
3
备注:
* 1 ≤ n ≤ 8
*
* 1 ≤ ai, bi ≤ n
* The number of test cases does not exceed 50.
题意:找出G2子图与G1同构的个数
思路:通过对点的全排列,然后判断在G2中的边,在当时的排列中G1的两点是否包含边,就是要G1的边与G2去对应,然后就会有多种情况,然后可以通过hash去重。
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=100000+10;
const double eps=1e-15;
const int mod=1e9+7;
typedef pair<int,int> P;
int n,m1,m2,num,a1,b1,z,k,map1[9][9],a[9];
set<ll>s;
P map2[30];
void init()
{
memset(map1,0,sizeof(map1));
for(int i=1;i<=n;i++)
a[i]=i;
s.clear();
}
int main()
{
while(~scanf("%d%d%d",&n,&m1,&m2))
{
init();
int u,v;
for(int i=1;i<=m1;i++)
{
scanf("%d%d",&u,&v);
map1[u][v]=map1[v][u]=1;
}
for(int i=1;i<=m2;i++)
scanf("%d%d",&map2[i].first,&map2[i].second);
do{
int cnt=0;
ll hashs=0;
for(int i=1;i<=m2;i++)//枚举G2的边
if(map1[a[map2[i].first]][a[map2[i].second]])//在G2中的边与G1边的对应
{
hashs=hashs*101+i;
hashs%=mod;
cnt++;
}
if(cnt==m1)
s.insert(hashs);
}while(next_permutation(a+1,a+n+1));//对G1进行全排列
printf("%d\n",s.size());
}
}