2015多校联合训练赛 hdu 5305 Friends 2015 Multi-University Training Contest 2 枚举+剪枝

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 249    Accepted Submission(s): 103

Problem Description

There are  n  people and  m  pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these  n  people wants to have the same number of online and offline friends (i.e. If one person has  x  onine friends, he or she must have  x  offline friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements. 

 

Input

The first line of the input is a single integer  T (T=100) , indicating the number of testcases. 

For each testcase, the first line contains two integers  n (1≤n≤8)  and  m (0≤m≤n(n−1)2) , indicating the number of people and the number of pairs of friends, respectively. Each of the next  m  lines contains two numbers  x  and  y , which mean  x  and  y  are friends. It is guaranteed that  x≠y  and every friend relationship will appear at most once. 

 

Output

For each testcase, print one number indicating the answer.

 

Sample Input

   
   
   
   

    
    
    
    2
3 3
1 2
2 3
3 1
4 4
1 2
2 3
3 4
4 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
2
   
   
   
   

 

Source

2015 Multi-University Training Contest 2

题意：

 给一个图，不存在重边。

任意两个人（结点）有线上或者线下的关系，但是每个人线上和线下关系的朋友必须一样多。

求有几种方案。

=====等效于对图的边黑白染色。每个点的黑色边和白色边要一样。

分析：

   特判如果存在边的度数为奇数的，直接输出0.

    只有八个人。实际上最复杂的情况是每个人的边数都是6条。那么枚举每个人，把他们的边染色。选出一半的边染成黑色。

   分析最复杂的情况C(3,6)*C(3,6)*C(2,4)*C(2,4)*C(1,2)*C(1,2)

为什么后面的C边数少了呢？因为之前枚举点的时候已经处理了这些边。所以后面就不用处理了。

剪枝：如果一个点的黑色边数 > 入度的一半，结束这次搜索

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
vector <int >head[10];
int online[10];
int in[10];
int n;
int dfs(int u);
int dfs2(int u,int p,int num){
    if(num == 0){
        dfs(u+1);
        return 0;
    }
    for(;p < head[u].size() && num+p <= head[u].size(); p++){
        int v = head[u][p];
        online[v]++;
        if(online[v] * 2 <= in[v]){
            dfs2(u,p+1,num-1);
        }
        online[v]--;
    }
    return 0;
}
int ans;
int dfs(int u){
    if(u == n+1){
        ans++;
        return 0;
    }
    int need = in[u] / 2 - online[u];
    dfs2(u,0,need);
}
int main(){

    int out[10];
    int t;
    scanf("%d",&t);
    while(t--){
        int m;
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n; i++)
            head[i].clear();
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        int u,v;
        for(int i = 0;i < m; i++){
            scanf("%d%d",&u,&v);
            if(u > v) swap(u,v);
            head[u].push_back(v);
            in[u]++,in[v]++;
        }
        int flag = 1;
        for(int i = 1;i <= n; i++){
            if(in[i] % 2 == 1) flag = 0;
        }
        if(flag){
            memset(online,0,sizeof(online));
            ans = 0;
            dfs(1);
            printf("%d\n",ans);
        }
        else printf("0\n");
    }
    return 0;
}