HDU1518:Square

Posted 2020-07-31 vCoders

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12571    Accepted Submission(s): 4008

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

 

思路:先排序，再dfs，详细见代码。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=25;
int n,sum,l;
int sticks[MAXN];
int vis[MAXN];
bool dfs(int dep,int len,int start,int num)//dep:dfs深度,len:目前sticks组成的长度,start:开始深搜的位置,num:组成的边数 
{
    if(dep==n)
    {
        if(num==4)
            return true;
        else
            return false;
    }
    for(int i=start;i<n;i++)
    {
        if(!vis[i]&&len+sticks[i]<=l)    
        {
            vis[i]=1;
            if(len+sticks[i]==l)
            {
                if(dfs(dep+1,0,0,num+1))//组成一个边从头开始搜 
                    return true;
            }
            else
            {
                if(dfs(dep+1,len+sticks[i],i+1,num))//继续向前搜 
                    return true;
            }
            vis[i]=0;
        }
    }
    return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(vis,0,sizeof(vis));
        sum=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&sticks[i]);
            sum+=sticks[i];
        }
        if(sum%4!=0)
        {
            printf("no\n");
        }
        else
        {
            sort(sticks,sticks+n);//排序 
            l=sum/4;    
            if(dfs(0,0,0,0))
                printf("yes\n");
            else
                printf("no\n");
        }
    }
    
    return 0;
}