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hdu 1518 Square

2024-07-14 04:13:31   232人阅读

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8260    Accepted Submission(s): 2687


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
 

<span style="font-size:18px;">#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define M 25
int a[M],vis[M];
int n,m,sum;

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

int dfs(int ans,int cur,int num)  //ans表示当前位置 cur表示当前权值 num表示边数
{  
	int i;
	if(num==4) return 1;
	for(i=ans;i<n;i++)
	{
		if(vis[i]) continue;
		if(cur+a[i]==m)  // 满足边长
		{
			vis[i]=1;
			if(dfs(0,0,num+1))  // 找下一条边  
				return 1;
			vis[i]=0;
		}
		else if(cur+a[i]<m)   //不满足边长
		{
			vis[i]=1;
			if(dfs(i,cur+a[i],num))
				return 1;
			vis[i]=0;
		}
	}
	return 0;
}


int main ()
{
	int i,j,t;
	cin>>t;
	while(t--)
	{
		cin>>n;
		sum=0;
		for(i=0;i<n;i++)
		{
			cin>>a[i];
			sum+=a[i];
		}
		if(sum%4!=0)
			printf("no\n");
		else
		{
			m=sum/4;    // 边长
			sort(a,a+n,cmp);
			if(m<a[0])    // 最大的数大于边长肯定不满足
				printf("no\n");
			else
			{  
				memset(vis,0,sizeof(vis));
				if(dfs(0,0,0))
					printf("yes\n");
				else
					printf("no\n");
			}
		}
	}
	return 0;
}</span>


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