hdu1518（Square）深搜+剪枝

点击打开杭电1518

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
   
   
   
   

题意：判断所给的几个数能否组成一个正四边形

思路：从题意可以看出，要成为正四边形，那么必须所给的数之和能别4整除，而且所给的数中最小的数不能大于正四边形的边长。然后通过深搜遍历每个数，直到组合成边长的长度为止，最后通过统计组成边长的个数，判断是否满足条件

import java.util.Scanner;

public class P1518 {
	public static int m,edglen;
	public static int[] visit;
	public static int[] a;
	public static boolean flag;
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		while(n-->0){
			m=sc.nextInt();
			a=new int[m];
			visit=new int[m];
			int sum=0;
			flag=false;
			for(int i=0;i<m;i++){
				a[i]=sc.nextInt();
				sum+=a[i];
			}
			if(sum%4!=0){//剪枝，总和不能被4整除，则一定不能组成
				System.out.println("no");
				continue;
			}
			edglen=sum/4;
			sort();
			if(edglen<a[0]){//最小的都大于边长，减去
				System.out.println("no");
				continue;
			}
			DFS(0,a[0],0);
			if(flag){
				System.out.println("yes");
			}else{
				System.out.println("no");
			}
		}
	}
	private static void DFS(int edglenNumb, int edg, int i) {
		int k=i;
		visit[i]=1;
		if(edg==edglen){
			edglenNumb++;
			k=0;
			edg=0;
		}
		if(edglenNumb==3){
			flag=true;
			return;
		}
		for(int j=k;j<m;j++){
			if(visit[j]==0 && edg+a[j]<=edglen){
				DFS(edglenNumb,a[j]+edg,j);
				if(flag){
					return ;
				}
			}
		}
		visit[i]=0;
	}
	private static void sort() {
		for(int i=0;i<m-1;i++){
			for(int j=0;j<m-1-i;j++){
				if(a[i]>a[j]){
					int t=a[i];a[i]=a[j];a[j]=t;
				}
			}
		}
	}
	
}