学步园

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

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.

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

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

yes
no
yes
#include<stdio.h>
#include<string.h>
int vist[25],a[25],flog,m,edglen;

void DFS(int edgnamber,int edgsum ,int i)
{
    int j,t=i;
    vist[i]=1;
    if(edgsum==edglen)//如果相等，则成立了一个边，边数就加1，搜组成下一个边
    {
        edgnamber++;edgsum=0;t=0;//t是下一个边搜的起点，edgsum是边的长度
    }
    if(edgnamber==3)//等于3说明成立了三个边，那么下一边也就不用搜了，直接反回
    {
        flog=1;
        return ;
    }
    for(j=t;j<m;j++)
    if(!vist[j]&&edgsum+a[j]<=edglen)//动态地加边长
    {
        DFS(edgnamber,edgsum+a[j],j);
        if(flog)
        return ;
    }
    vist[i]=0;
}

int main()
{
    int t,i,sum;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        if(sum%4!=0||m<4)//剪枝
        {
            printf("no\n");
            continue;
        }
        flog=0; edglen=sum/4;
        memset(vist,0,sizeof(vist));
        DFS(0,a[0],0);
        if(flog)
        printf("yes\n");
        else
        printf("no\n");
    }
}