POJ 2362(Square-搜索剪枝1-相对顺序)

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内容目录
Language:
Square
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 17066   Accepted: 5878

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

Waterloo local 2002.09.21

《搜索是怎样剪枝的-1》

1.只要找到3条边。

2.从大到小顺序找。

3.每次搜边时要确定边的相对顺序唯一（直接从TLE→秒）


#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<cstring>
#include<iostream>
using namespace std;
#define MAXM (20+10)
int tt,n,a[MAXM],cnt,len;
bool b[MAXM],flag;
bool dfs(int l,int x,int pre)
{
//	cout<<l<<' '<<x<<' '<<kth<<endl;
	if (x==len) {l++;x=0;pre=n-1;}
	if (l==4)
	{
		return 1;
	}
	for(int i=pre-1;i;i--)
		if (!b[i]&&x+a[i]<=len)
		{
			b[i]=1;
			if (dfs(l,x+a[i],i)) return 1;
			b[i]=0;
		//	if (!x) return 0;
		}
	return 0;
}
int main()
{
	scanf("%d",&tt);
	while (tt--)
	{
		cnt=0;
		scanf("%d",&n);
		for (int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			cnt+=a[i];b[i]=0;
		}
		sort(a+1,a+1+n);
		if (n<4||cnt%4||a[n]>cnt/4)
		{
			printf("non");continue;
		}
		b[n]=1;len=cnt/4;
		if (dfs(1,a[n],n))
		{
			printf("yesn");
		}
		else printf("non");
	}
	return 0;
}