牛客网多校训练第一场 D

链接：

https://www.nowcoder.com/acm/contest/139/D

题意：

两个无向简单图都有n（1≤n≤8）个顶点，图G1有m1条边，图G2有m2条边，问G2有多少个子图与G1同构。

分析：

枚举所有的映射方案，再判断合法性，把每一合法映射所用到的边状态压缩一下，装到集合里去重即可。

代码：

import java.io.*;
import java.util.*;

public class Main {
    Scanner cin = new Scanner(new BufferedInputStream(System.in));
    final int UP = 8 + 5;
    int n, map[] = new int[99], cor[] = new int[UP];
    boolean G[][] = new boolean[UP][UP], G2[][] = new boolean[UP][UP];
    boolean opt[] = new boolean[UP];
    HashSet<Integer> set = new HashSet<Integer>();
    
    void init() {
        for(int r = 0; r < UP; r++)
            for(int c = 0; c < UP; c++) G[r][c] = G2[r][c] = false;
    }
    
    void dfs(int d) {
        if(d == n + 1) {
            int code = 0;
            for(int f = 1; f <= 8; f++) {
                for(int b = 1; b <= 8; b++) {
                    if(!G[f][b]) continue;
                    if(!G2[cor[f]][cor[b]]) return;
                    code |= 1 << map[cor[f]*10+cor[b]];
                }
            }
            set.add(code);
            return;
        }
        for(int i = 1; i <= n; i++) {
            if(opt[i]) continue;
            opt[i] = true;
            cor[d] = i;
            dfs(d+1);
            opt[i] = false;
        }
    }
    
    void MAIN() {
        while(cin.hasNext()) {
            init();
            n = cin.nextInt();
            int m = cin.nextInt();
            int m2 = cin.nextInt();
            for(int f, b, i = 0; i < m; i++) {
                f = cin.nextInt();
                b = cin.nextInt();
                G[f][b] = G[b][f] = true;
            }
            for(int f, b, i = 0; i < m2; i++) {
                f = cin.nextInt();
                b = cin.nextInt();
                G2[f][b] = G2[b][f] = true;
                map[f*10+b] = map[b*10+f] = i;
            }
            
            set.clear();
            dfs(1);
            System.out.println(set.size());
        }
    }
    
    public static void main(String args[]) { new Main().MAIN(); }
}