poj 2365squre 正方形

Square

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25421		Accepted: 8669

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

解题说明：重要的剪纸：

1，当前拼的边长度+最小的长度>实际边长度。

2，上一个长度不行，下次和该长度相同就不要继续搜了。

3，先排个序，减少搜索树的高度。

代码：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<functional>
using namespace std;
int n,mn,mx;
int sum;
int a[25];
int mark[25];
int last=0;
int dfs(int x,int ans,int tot){
	if(ans==3)return 1;
	if(tot==sum){
	   ans++;
	   x=1;
	   tot=0;
	}
	if(tot+mn>sum)return 0;
	for(int i=x+1;i<=n;i++){
		if(!mark[i]){
			if(last==a[i])continue;
			mark[i]=1;
			if(dfs(i,ans,tot+a[i]))return 1;
			mark[i]=0;
			last=a[i];
			
		}
	}
	return 0;
}
int main(){
	int T;scanf("%d",&T);
	while(T--){
		memset(a,0,sizeof(a));
		memset(mark,0,sizeof(mark));
		scanf("%d",&n);
		sum=0;
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		    sum+=a[i];	
	 	}
		sort(a+1,a+1+n,greater<int>());
		mx=a[1];mn=a[n];
		if(sum%4||sum/4<mx){
			printf("no\n");
			continue; 
		}
		sum/=4;
		mark[1]=1;
		if(dfs(1,0,a[1]))printf("yes\n");
		else printf("no\n");
	}
	return 0;
}