poj 2362 Square

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

这个题可以说是poj1011的减弱版，只需要查一个值，不过在dfs时候的剪枝基本上是一模一样的

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;

bool cmp(int x,int y)
{
    return x>y;
}

int n,a[21];
bool vis[21];

bool dfs(int nowlen,int Len,int x,int num)
{
    if (num==n) return true;
    int temp=0;
    for (int i=x;i<n;i++)
    {
        if (vis[i] || a[i]==temp) continue;
        vis[i]=true;
        if (nowlen+a[i]<Len)
        {
            if (dfs(nowlen+a[i],Len,i+1,num+1)) return true;
            else temp=a[i];
        }
        if (nowlen+a[i]==Len)
        {
            if (dfs(0,Len,0,num+1)) return true;
            else temp=a[i];
        }
        vis[i]=false;
        if (nowlen==0) break;
    }
    return false;
}

int main()
{
    int t,sum;
    scanf("%d",&t);
    while (t--)
    {
        sum=0;
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for (int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sort(a,a+n,cmp);
        if (sum%4==0)
        {
            if (dfs(0,sum/4,0,0))
                printf("yes
");
            else printf("no
");
        }
        else printf("no
");
    }
    return 0;
}

作者 chensunrise