Square（DFS）

最新推荐文章于 2022-05-15 15:28:02 发布

原创最新推荐文章于 2022-05-15 15:28:02 发布 · 401 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#dfs #c

dfs 专栏收录该内容

5 篇文章

订阅专栏

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

解题思路

n个木棒能否构成正方形且都要用上，按边来深搜，

当返回到最初的dfs中，for循环结束未能return 1;则返回值0到主函数中；

count是当前层的count，如当从第二条边回溯到第一条边时，此时count=0；

AC代码

#include<stdio.h>
#include<string.h>

int a[21];
bool used[21];
int n,m,sum;

int dfs(int d, int w, int count)
{
	if(d == sum)                                          //d==sum时返回值1到上一层，进行return 1;再返回到上一层，
	{
		w=0;   d=0; count++;
		if(count == 3)
			return 1;
	}

	for(int i=w; i<m; i++)
		if(!used[i])
		{
			used[i] = 1;
			if((d+a[i] <= sum ) && dfs(d+a[i], i+1, count))  
              		    return 1;
			used[i] = 0;
		}

	return 0;                                          //当前边不能构成，返回上一层
}

int main()
{
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d",&m);
		sum=0;
		memset(used, 0, sizeof(used));
		for(int i=0; i<m; i++)
		{
			scanf("%d",&a[i]);
			sum += a[i];
		}

		int flag=0;
		if(sum%4 == 0)                                  //构成正方形的先行条件
		{
			sum /= 4;  
			flag=dfs(0, 0, 0);
		}

		if(flag)
			printf("yes\n");
		else
			printf("no\n");
	}

	return 0;
}