HDU 1518 Square

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6187    Accepted Submission(s): 2013

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

 

Source

University of Waterloo Local Contest 2002.09.21

 

Recommend

LL

 

思路：DFS

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int map[100];
int n,m;
int used[100];
int sum;
int side;
int flag;
int mmm;
void DFS(int now,int cnt,int k)
{
    if(now == side)
    {
        cnt ++;
        now = 0;
        k = 0;
    }
    if(cnt == 4)
    {
        flag = 1;
        return ;
    }
    for(int i = k;i <= m;i ++)
    {
        if(!used[i])
        {
            used[i] = 1;
            if(now + map[i] <= side)
            {
                DFS(now + map[i],cnt,i + 1);
                if(flag)
                   return ;
            }
            used[i] = 0;
        }
    }
}
int main()
{
    scanf("%d",&n);
    while(n --)
    {
        scanf("%d",&m);
        sum = 0;
        mmm = 0;
        flag = 0;
        for(int i = 1;i <= m;i ++)
        {
              scanf("%d",&map[i]);
              if(mmm < map[i])
                  mmm = map[i];
              sum += map[i];
        }
        if(sum % 4 != 0 || mmm > sum / 4)
        {
            flag = 0;
        }
        else
        {
            side = sum / 4;
            memset(used,0,sizeof(used));
            DFS(0,0,0);
        }
        if(flag == 1)
           printf("yes ");
        else
           printf("no ");
    }
    return 0;
}