Square POJ - 2362

Square

 POJ - 2362 

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

题解+解析：

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#define INF 0x7f7f7f7f
typedef long long ll;
using namespace std;
int n;
int m[25];
bool visit[25];
int sum, mid;
int flag;
void dfs(int num, int len, int s)//num为凑好的边的个数，len为现在在凑的木棍长度，s代表起点位置
{
	if (flag)//
	{
		return;
	}
	if (num == 4)
	{
		flag = true;
		return;
	}
	if (len == mid)
	{
		dfs(num+1, 0, 0);//符合边长，深搜下一条边
	}
	for (int i = s; i < n; i++)
	{
		if (!visit[i]&&len+m[i]<=mid)//不搜重复和不符合条件的
		{
			visit[i] = true;
			dfs(num, len+m[i], i+1);//符合条件就搜一下
			visit[i] = false;//回溯
		}
	}
}
int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		sum = 0;
		flag = false;
		memset(visit, 0, sizeof(visit));
		scanf("%d", &n);
		for (int i =0; i < n; i++)
		{
			scanf("%d",&m[i]);
			sum += m[i];
		}
		sort(m,m+n);//排序
		mid = sum/4;
		if(mid*4!=sum)//如果mid不是整数，直接输出no
		{
			printf("no\n");
			continue;
		}
		if (m[n-1]>mid)//如果最长的棍大于边长，直接输出no
		{
			printf("no\n");
			continue;
		}
		dfs(0,0,0);
		if (flag)
		{
			printf("yes\n");
		}
		else
		{
			printf("no\n");
		}
	}
	return 0;
}