Two Graphs 牛客网暑期ACM多校训练营（第一场）D 图论基础知识全排列

链接：https://www.nowcoder.com/acm/contest/139/D
来源：牛客网

Two undirected simple graphs $G_1 = \langle V, E_1 \rangle$ and $G_2 = \langle V, E_2 \rangle$ where $V = \{1, 2, \dots, n\}$ are isomorphic when there exists a bijection $\phi$ on V satisfying $\{\phi(x), \phi(y)\} \in E_1$ if and only if {x, y} ∈ E₂.
Given two graphs $G_1 = \langle V, E_1 \rangle$ and $G_2 = \langle V, E_2 \rangle$ , count the number of graphs $G = \langle V, E \rangle$ satisfying the following condition:
* $E \subseteq E_2$ .
* 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, m

₁

 and m

₂

 where |E

₁

| = m

₁

 and |E

₂

| = m

₂

.
The i-th of the following m

₁

 lines contains 2 integers a

 and b

 which denote {a

, b

} ∈ E

₁

.
The i-th of the last m

₂

 lines contains 2 integers a

 and b

 which denote {a

, b

} ∈ E

₂

输出描述:

For each test case, print an integer which denotes the result.

输入例子:

输出例子:

2

3

-->

示例1

输入

复制

输出

复制

2

3

备注:

* 1 ≤ n ≤ 8
*

$1 \leq m_1 \leq m_2 \leq \frac{n(n - 1)}{2}$

* 1 ≤ a

, b

 ≤ n
* The number of test cases does not exceed 50.

求图G1有多少个子图和图G2是同构图
枚举图G1的全排列子图，然后判断每一个子图是否和图G2是同构图
判断原理：
同构要求边，点以及点的结构都相同。
我们先存好图G1中每种边点结构的关系，用一个vis数组存下每条边
然后在全排列判断每个子图的时候判断这个子图是否和图G2同构，同构就用个set保存下这种情况（每种情况用散列哈希求出一个独一无二的值）
最后set的size就是结果

#include <map>

#include <set>

#include <stack>

#include <cmath>

#include <queue>

#include <cstdio>

#include <vector>

#include <string>

#include <cstring>

#include <iostream>

#include <algorithm>

#define debug(a) cout << #a << " " << a << endl

using namespace std;

const int maxn = 1e1;

const int mod = 1e9 + 7;

typedef long long ll;

ll vis[maxn][maxn], f[maxn];

pair<ll,ll> p[maxn*10];

int main() {

    ll n, m1, m2;

    while( cin >> n >> m1 >> m2 ) {

        memset( vis, 0, sizeof(vis) );

        for( ll i = 1; i <= n; i ++ ) {

            f[i] = i;

        }

        for( ll i = 1, a, b; i <= m1; i ++ ) {

            cin >> a >> b;

            vis[a][b] = vis[b][a] = 1;

        }

        for( ll i = 1; i <= m2; i ++ ) {

            cin >> p[i].first >> p[i].second;

        }

        set<ll> s;

        do{

            ll ha = 0, cnt = 0;

            for( ll i = 1; i <= m2; i ++ ) {

                if( vis[f[p[i].first]][f[p[i].second]] ) {  //判断边点结构

                    ha = ha*133 + i; //散列求点，确保每种情况得到的点不一样

                    ha %= mod;

                    cnt ++;   //判断边

                }

            }

            if( cnt == m1 ) {

                s.insert(ha);

            }

        } while( next_permutation(f+1,f+n+1) ); //全排列图G1

        cout << s.size() << endl;

    }

    return 0;

}