Square

最新推荐文章于 2020-01-07 21:33:25 发布

k_Tao

最新推荐文章于 2020-01-07 21:33:25 发布

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分类专栏：搜索文章标签： acm

本文链接：https://blog.youkuaiyun.com/qq_32175783/article/details/51413202

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Description

Problem C: Square

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

yesd



该题考察深度搜索和回溯，为了避免超时，在搜索之前进行剪枝。例如这题，题意：给出一些长度，看能否围成正方形，

我们可以先计算每个边的边长看是否是整数，因为长度不可分割，所以如果不是整数，就直接输出"no"，如果其中一个长度

大于平均长度，也不能构成正方形，输出"no"，然后再进行ｄｆｓ（０，０，０）搜索。







#include
 <cstring>
#include
 <iostream>
usingnamespacestd;
inta[21],
 vis[21];
intflag,
 sum, ave, n;
voiddfs(intx,inty,intp)
{
    inti;
    if(x==4)
    {
        flag=1;
        return;
    }
    if(y==ave)
    {
        dfs(x+1,0,0);//从0开始重新搜索
        if(flag)
        return;
    }
    for(i=p;
 i<n;i++)
    {
        if(!vis[i]&&y+a[i]<=ave)
        {
            vis[i]=1;
            dfs(x,a[i]+y,i+1);
            if(flag)
                return;
            vis[i]=0;
        }
    }
}
 
intmain()
{
    intT,
 i;
    while(cin>>T)
    {
        while(T--)
        {
            memset(vis,0,sizeof(vis));
            sum=0,flag=0;
            cin>>n;
            for(i=0;i<n;i++)
            {
                cin>>a[i];
                sum+=a[i];
            }
            if(sum%4!=0)
            {
                cout<<"no"<<endl;
                continue;
            }
            ave=sum/4;
            for(i=0;i<n;i++)
            {
                if(a[i]>ave)
                {
                    flag=1;
                    break;
                }
            }
            if(flag==1)
            {
                cout<<"no"<<endl;
                continue;
            }
            dfs(0,0,0);
            if(flag)
                cout<<"yes"<<endl;
            else
                cout<<"no"<<endl;
        }
    }
    return0;
}