牛客网暑期ACM多校训练营（第一场）D Two Graphs

gtuif

于 2018-08-07 10:43:36 发布

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分类专栏：图的同构

本文链接：https://blog.youkuaiyun.com/gtuif/article/details/81476854

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本文介绍了一种解决图同构问题的方法，通过全排列遍历节点并检查边的存在性来判断两个图是否同构，同时考虑了自同构的情况。

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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

输出

复制

2
3

备注:

* 1 ≤ n ≤ 8
* 
* 1 ≤ ai, bi ≤ n
* The number of test cases does not exceed 50.

题意：问B的子图有多少个和A同构；

思路：首先明白什么是同构，对于图的话，简单来说，节点编号变但是图形没变，可以称这两个图同构；那么我们直接对节点进行全排列，然后遍历m1条边，看两个点构成的边是否在B中存在，当然这个过程中还会有A的自同构（边的两个节点编号没变成别的数，比如顶点互换），除去就好了；

下面附上代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#define long long LL
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define lson o<<1
#define rson o<<1|1
using namespace std;
int g1[100][100];
int g2[100][100]; 
int u[100], v[100];
int a[10];
int main()
{
	int n, m1, m2;
	while(cin >> n >> m1 >> m2)
	{
		int x, y;
		memset(a, 0, sizeof(a));
		memset(g1, 0, sizeof(g1));
		memset(g2, 0, sizeof(g2));
		for(int i = 0; i < m1; i++)
		{
			scanf("%d %d",&u[i], &v[i]);
			g1[u[i]][v[i]] = g1[v[i]][u[i]] = 1;
		}
		for(int i = 0; i < m2; i++)
		{
			scanf("%d %d", &x, &y);
			g2[x][y] = g2[y][x] = 1;
		}
		for(int i = 1; i <= n; i++)
			a[i] = i;
		int ams = 0;int ans = 0;
		do{
			int ans1 = 1, ans2 = 1;
			for(int i = 0; i < m1; i++)
			{
				x = a[u[i]];
				y = a[v[i]];
				if(!g1[x][y])//没有自同构 
					ans1 = 0;
				if(!g2[x][y])//不同构 
					ans2 = 0;	
			}		
			ans += ans1;
			ams += ans2;
		}while(next_permutation(a + 1, a + n + 1));
		printf("%d\n", ams/ans);
	}
    return 0;
}