（HDU - 1518）Square

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（HDU - 1518）Square

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

Total Submission(s): 16247 Accepted Submission(s): 5120

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

题目大意：给出n根木棒，问这n根木棒能都构成正方形，所有木棒都要用到。

思路：dfs+剪枝。dfs需要回溯，因为标记后，如果那种方法不行，回溯才能使下次能搜到这些之前被标记过但这次还需要再搜的边。信仰剪枝，大胆暴力出奇迹….当时没怎么想，就是一发暴力，结果900ms爬过…

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

const int maxn=25;
int a[maxn];
bool vis[maxn];
int m,avelen;

bool dfs(int cnt,int tmplen,int id)//cnt表示已经成功的边数，tmplen表示当前边的长度，id表示当前所用棍子编号 
{
    if(cnt==4) return true;
    if(tmplen==avelen) 
    {
        if(dfs(cnt+1,0,0)) return true; 
    }
    for(int i=id;i<m;i++)
        if(!vis[i]&&tmplen+a[i]<=avelen)
        {
            vis[i]=1;
            if(dfs(cnt,tmplen+a[i],i+1)) return true;
            vis[i]=0;//回溯 
        }
    return false;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&m);
        int sumlen=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d",a+i);
            sumlen+=a[i];
        }
        avelen=sumlen/4;
        sort(a,a+m);
        if(sumlen%4||a[m-1]>avelen)//不能被4整除或者最长的木棒比边长还长，则不能构成正方形 
        {
            printf("no\n");
            continue;
        }
        memset(vis,0,sizeof(vis));
        bool flag=dfs(0,0,0);
        if(flag) printf("yes\n");
        else printf("no\n");
    }
    return 0;
}