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，看可以构成正方形，这题的数据比較水，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表示下次从哪根開始拼，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
");continue;
		}
		liang=sum/4;
		sort(a+1,a+1+n,cmp);
		memset(vis,0,sizeof(vis));
		if(dfs(0,n,0))printf("yes
");
		else printf("no
");
	}
	return 0;
}