2015 多校联赛 ——HDU5305（搜索）

Friends

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 163 Accepted Submission(s): 61

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.

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

求：朋友关系可离线可在线，要求每个人的离线朋友数等于在线朋友数，总共有多少种方法

先对入度判断，奇数直接不可能，然后再进行搜索

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <vector>
using namespace std;

struct node
{
int u,v;
} pnode[50];

int in[10];
int c[10],c2[10];
int ans,n,m;
int num,tem;

void dfs(int u)
{
if(u == m+1)
{
for(int i = 1; i <= n; i++)
if(c[i] != c2[i])
return;
ans++;
return ;
}
int a = pnode[u].u;
int b = pnode[u].v;
if(c[a] < in[a]/2 && c[b] < in[b]/2)
{
c[a] ++;
c[b] ++;
dfs(u+1);
c[a]--;
c[b]--;
}
if(c2[a] < in[a]/2 && c2[b] < in[b]/2)
{
c2[a] ++;
c2[b] ++;
dfs(u+1);
c2[a]--;
c2[b]--;
}
return;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int flag = 0;
scanf("%d%d",&n,&m);
memset(in,0,sizeof(in));
for(int i = 1; i <= m; i++)
{
scanf("%d%d",&pnode[i].u,&pnode[i].v);
in[pnode[i].u]++;
in[pnode[i].v]++;
}
memset(c,0,sizeof(c));
memset(c2,0,sizeof(c2));
for(int i = 1; i <= n; i++)
{
if(in[i] % 2 == 1)
{
flag = 1;
break;
}
}
if(flag){
printf("0\n");
continue;
}
ans = 0;
dfs(1);

printf("%d\n",ans);

}
return 0;
}