Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Source

University of Waterloo Local Contest 2002.09.21

#include"stdio.h"
#include"string.h"
#include"queue"
#include"vector"
#include"algorithm"
using namespace std;
#define N 25
#define max(a,b) (a>b?a:b)
int a[N],n,t,mark[N],sum;
bool cmp(int a,int b)
{
    return a>b;
}
int dfs(int index,int len,int rest)
{
    int i;
    if(rest==t)
        return 1;
    for(i=index;i<n;i++)
    {
        if(!mark[i]&&a[i]<=len)
        {
            mark[i]=1;
            if(len==a[i])
            {
                if(dfs(0,t,rest-a[i]))
                    return 1;
            }
            else if(dfs(i+1,len-a[i],rest-a[i]))
                return 1;
            mark[i]=0;  //若当前值不能满足条件，进行回溯
            if(rest==sum)       //若满足此条件，则该条边不能加入任一条边
                return 0;
            if(len==t)   //同上
                return 0;
            while(a[i]==a[i+1])     //这条边不行，则和它同长的边也不行
                i++;
        }
    }
    return 0;
}
int main()
{
    int T,i;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n,cmp);
        if(sum%4)
            printf("no\n");
        else
        {
            memset(mark,0,sizeof(mark));
            if(dfs(0,t=sum/4,sum))
                printf("yes\n");
            else
                printf("no\n");
        }
    }
    return 0;
}

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