hdoj5305【dfs】

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2078    Accepted Submission(s): 1027

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

23 31 22 33 14 41 22 33 44 1

 

Sample Output

02

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 2

 

题意：n个人m对朋友关系，每个人可以选择他的朋友为网络朋友或现实朋友且这两种朋友的个数必须相等，问有多少种方法满足条件。

解题思路：按照所给的朋友关系要么这对关系为网路朋友要么为现实朋友dfs即可

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<list>#include<queue>#include<vector>using namespace std;const int maxn=100;int n,m;long long ans=0;struct Node{int a,b;}A[70];int ol[maxn],of[maxn],vis[10];;bool judge(){for(int i=1;i<=m;++i){if(of[A[i].a]!=ol[A[i].a]||of[A[i].b]!=ol[A[i].b])return false;}return true;}void dfs(int k){if(k==m+1){//if(judge())ans++;return ;}ol[A[k].a]++;ol[A[k].b]++;if(ol[A[k].a]<=vis[A[k].a]/2&&ol[A[k].b]<=vis[A[k].b]/2)dfs(k+1);ol[A[k].a]--;ol[A[k].b]--;of[A[k].a]++;of[A[k].b]++;if(of[A[k].a]<=vis[A[k].a]/2&&of[A[k].b]<=vis[A[k].b]/2)dfs(k+1);of[A[k].a]--;of[A[k].b]--;}int main(){int t,i,j,k;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);memset(ol,0,sizeof(ol));memset(of,0,sizeof(of));memset(vis,0,sizeof(vis));for(i=1;i<=m;++i){scanf("%d%d",&A[i].a,&A[i].b);vis[A[i].a]++;vis[A[i].b]++;}for(i=1;i<=n;++i){if(vis[i]&1)break;}if(i<=n){printf("0\n");continue;}ans=0;dfs(1);printf("%lld\n",ans);}return 0;}