HDU 1518 Square DFS

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Square

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13645 Accepted Submission(s): 4324

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

Source

University of Waterloo Local Contest 2002.09.21

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题意大概就是给你几个数代表棍的长度，让你判断他能不能构成正方形（标准的），我用的是深度搜索，一遍遍搜索，用now作为判断搜索的第几次，然后超时了，上网上看了看大神的代码，他加了个index，用来降低时间复杂度，

index的大体意思是：每一遍扫描代表扫出来一条边，所以只能往前走，不会拐过来在用以前的数值的，所以扫描过的没必要在扫描一遍了。还有flag全局变量判断及在dfs中的剪纸，我这里简单说一下，第一个if是因为判断到第五条边了，然后四条就够，所以结束，把flag赋值为1。dfs中的第二个if里面有个判断flag的，就是第五条边都出来了，下面的就不用再去计算了。dfs for循环里面的flag判断也是类似。这样剪完之后就可以过了~~~

#include <stdio.h>

int save[25];
int vis[25];
int m,len,flag;

void dfs(int now,int l,int index)
{
	int i;
	
	if(now==5)
	{
		flag=1;
		return ;
	}
	if(l==len)
	{
		dfs(now+1,0,0);
		
		if(flag)
			return ;
	}
	for(i=index;i<m;i++)
	{
		if(!vis[i]&&(save[i]+l)<=len)
		{
			vis[i]=1;
			dfs(now,save[i]+l,i+1);
			if(flag)
				return ;
				
			vis[i]=0;
		}
	}
}

int main()
{
	int t,i,j,sum;
	
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&m);
		sum=0;
		for(i=0;i<m;i++)
		{
			scanf("%d",&save[i]);
			sum+=save[i];
		}
		if(sum%4)
		{
			printf("no\n");
			continue;
		}
		len=sum/4;
		for(i=0;i<m;i++)
		{
			if(len < save[i])
				break;
		}
		if(i<m)
		{
			printf("no\n");//如果有一个比边长大，则不符合 
			continue;
		}
		for(i=0;i<25;i++)
		{
			vis[i]=0;
		}
		flag=0;
		dfs(1,0,0);
		if(flag)
		{
			printf("yes\n");
		}
		else
		{
			printf("no\n");	
		}
		
	}
	
	return 0;
}