HDU1518 Square (DFS)

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14057    Accepted Submission(s): 4439

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

 

思路：一道基础的DFS题目，注意要剪枝，在进行搜索时，如果某一条边还没有搜索完毕，则下次搜索应该从数组a的下一个元素开始，因为在这之前的所有元素都已经搜索过了，再搜索就重复了。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

int a[25],vis[25],flag;

void dfs(int k,int cnt, int cur,int aver, int m){
	if(cnt == 3){
		flag = 1;
	}	
	if(flag)
		return;
		
	for(int i = k; i < m; i ++){
		if(!vis[i]){
			if(cur + a[i] < aver){
				vis[i] = 1;
				dfs(i + 1,cnt,cur+a[i],aver,m);
				vis[i] = 0;
			}
			else if(cur + a[i] == aver){
				vis[i] = 1;
				dfs(0,cnt+1,0,aver,m);
				vis[i] = 0;
			}
		}
	}
}

int main(){
	int n,m;
	scanf("%d",&n);
	while(n --){
		scanf("%d",&m);
		int sum = 0;
		for(int i = 0; i < m; i ++){
			scanf("%d",&a[i]);
			sum += a[i];
		}
		
		if(sum % 4){
			printf("no\n");
			continue;
		}
		flag = 0;
		for(int i = 0; i < m; i ++){
			if(a[i] > sum / 4){
				flag = 1;
				break;
			}
		}
		if(flag){
			printf("no\n");
			continue;
		}
		flag = 0;
		memset(vis,0,sizeof(vis));
		dfs(0,0,0,sum/4,m);	
		if(flag)
			printf("yes\n");
		else
			printf("no\n");
		
	}
	return 0;
}