poj2362 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

这题题意是给你一些边，看能够构成正方形，这题的数据比较水，为后面的poj1011埋下伏笔。

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<string>

#include<algorithm>

using namespace std;

int vis[30],liang,a[30],n;//liang表示每条边的长

bool cmp(int a,int b){

	return a<b;

}

int dfs(int x,int pos,int len)//x表示已经拼了几根，pos表示下次从哪根开始拼，len表示当前拼的这根已经拼了多少长度

{

	int i,j;

	if(x==3)return 1;

	for(i=pos;i>=1;i--){

		if(!vis[i]){

			if(a[i]+len<liang){

				vis[i]=1;

				if(dfs(x,i-1,len+a[i]))

				return 1;

				vis[i]=0;

			}

			else if(a[i]+len==liang){

				vis[i]=1;

				if(dfs(x+1,n,0))return 1;

				vis[i]=0;

			}

		}

	}

	return 0;

}

int main()

{

	int m,i,j,T,sum;

	scanf("%d",&T);

	while(T--)

	{

		scanf("%d",&n);

		sum=0;

		for(i=1;i<=n;i++){

			scanf("%d",&a[i]);

			sum+=a[i];

		}

		if(sum%4!=0){

			printf("no\n");continue;

		}

		liang=sum/4;

		sort(a+1,a+1+n,cmp);

		memset(vis,0,sizeof(vis));

		if(dfs(0,n,0))printf("yes\n");

		else printf("no\n");

	}

	return 0;

}

巴特西

poj2362 Square

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