HDU1518 Square

 

http://acm.hdu.edu.cn/showproblem.php?pid=1518
算法参考：http://blog.csdn.net/w00w12l/article/details/7865348

Problem Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

 

 

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

 

 

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

 

 

Sample Input

3 4 1 1 1 1 5 10 20 30 40 50 8 1 7 2 6 4 4 3 5

 

 

Sample Output

yes no yes

 

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;

int a[30];
bool vis[30];
int n,m,sum,len;
bool dfs(int cur,int time ,int k)
{
    if(time==3)  return true;
    for(int i=k; i>=1; i--)
    {
        if(!vis[i]){
            vis[i]=true;
            if(cur+a[i]==len){
                if(dfs(0,time+1,m)) return true;
            }
            else if(cur+a[i]<len){
                if(dfs(cur+a[i],time,i-1)) return true;     //关键是i-1..不然会一直超时的。。
                                                         
            }
            vis[i]=false;
        }
    }
    return false;
}
int main()
{
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m); sum=0;
        for(int i=1; i<=m; i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        memset(vis,0,sizeof(vis));
        sort(a,a+m);

        if(sum%4==0&&a[m]<=sum/4&&m>=4)
        {
            len=sum/4;
            if(dfs(0,0,m))  printf("yes\n");
            else            printf("no\n");
        }
        else  printf("no\n");

    }
    return 0;
}