HDU-1518: Square

搜索

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

题意：每组数据给出n根木棒，长短不一，问是否能恰好用这n根木棒拼出一个正方形，n根木棒要用完。

• 代码：

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[111],ave,l,n,ans;
int vis[111];
void dfs(int l,int count,int pos)//当前边长度，边数,当前搜索的位置
{
    if(l>ave||ans==1) return; //如果当前长度大于ave则不可行
    if(count==3)
    {
        ans=1;
        return;
    }
    if(l==ave)  //该边长度满足
    {
        count++;
        l=0; //重置l与pos，开始下一边搜索
        pos=n-1;
    }
    for(int i=pos;i>=0;i--)
    {
        if(vis[i]==0)  //如果该木棒未被使用过
        {
            vis[i]=1;
            dfs(l+a[i],count,i-1);
            vis[i]=0; //回溯
        }
    }
}
int main()
{
    int t,i,sum;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(vis,0,sizeof(vis));
        sum=0,l=1,ans=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n);
        ave=sum/4;
        if(n<=3||sum%4!=0||a[n-1]>ave) //预处理
            printf("no");
        else
        {
            dfs(0,0,n-1);
            if(ans==1)
                printf("yes");
            else
                printf("no");
        }
        printf("\n");
    }
    return 0;
}