DFS——hdu1518Square

一、题目回顾

题目链接：Square

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

二、解题思路

• DFS
• 剪枝

三、代码

//dfs+剪枝
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1e4+10;
#define INF 0x3f3f3f3f
int a[maxn];
int n,m,ave;						//边长
bool vis[maxn];
bool flag;

void dfs(int num,int len,int start)		//已凑完的边数，当前这条边的长度，从第几根木棍开始找
{
	 if(num==4){
	 	flag = 1;
	 	return;						//有4条边时,即可返回
	 }
	 if(len==ave){
	 	dfs(num+1,0,0);
	 	if(flag)
	 		return;
	 }
	 for(int i=start;i<m;i++){
	 	if(!vis[i] && a[i]+len<=ave){
	 		vis[i] = 1;
	 		dfs(num,a[i]+len,i+1);
	 		if(flag)	return;
	 		vis[i] = 0;					//剪枝，即第i个棍子加入不能形成正方形就不选此棍子
	 	}
	 }
}

int main()
{
	cin>>n;
	while(n--){
		cin>>m;
		int sum=0, maxlen=0;
		for(int i=0;i<m;i++){
			scanf("%d",&a[i]);
			sum += a[i];
			if(a[i]>maxlen)		maxlen = a[i];
		}
		ave = sum/4;
		if(sum%4!=0 || maxlen>ave){			//如果最长的边大于平均边长
			printf("no\n");
			continue;
		}
		sort(a,a+m);
		memset(vis,0,sizeof(vis));
		flag = 0;
		dfs(0,0,0);
		if(flag==1)	printf("yes\n");
		if(flag==0)	printf("no\n");
	}
	return 0;
}