hdu--1518--DFS(C&&java深搜之简单)

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5902    Accepted Submission(s): 1882

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

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

 

Sample Output

yesnoyes
#include<stdio.h>#include<stdlib.h>#include<string.h>#define max(a,b) ((a)>(b)?(a):(b))int a[25];int edge_len,m,tag[26];int search(int start,int s,int cnt){if(s==edge_len){return cnt==3?1:search(0,0,cnt+1);//return cnt==3?1:search(cnt+1,0,cnt+1); //当找到符合的一条边时，其中必然包含a[0],因为edge_len=a[0]+a[1]+...a[m-1]。//所以当搜索第二条边时，可以从cnt+1开始，也就是从a[1]开始  }else{int i;for(i=start; i<m; i++){if(!tag[i]&&s+a[i]<=edge_len){tag[i]=1;if(search(i+1,s+a[i],cnt))return 1;tag[i]=0;}}}return 0;}int main(){int n,i,sum;int maxn;scanf("%d",&n);while(n--){memset(tag,0,sizeof(tag));scanf("%d",&m);edge_len=sum=maxn=0;for(i=0; i<m; i++){scanf("%d",&a[i]);sum += a[i];maxn=max(maxn,a[i]);}if(sum%4 || maxn>sum>>2){printf("no\n");}else{edge_len=sum>>2;if(search(0,0,1)){printf("yes\n");}else{printf("no\n");}}}return 0;} 
 
 
 
java版：
 
import java.util.Arrays;import java.util.Comparator;import java.util.Scanner;public class hdu1518DFS{    private static boolean [] vis;    private static Integer [] array;    private static int edge;    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        int cases,n,sum;        cases = in.nextInt();        while(cases-->0){            n = in.nextInt();            sum = 0;            vis = new boolean[n];            array = new Integer[n];            for(int i = 0; i < array.length; ++i){                array[i] = in.nextInt();                sum += array[i];            }            edge = sum >>2;            if(sum%4 == 0 && edge >= array[0]){                Arrays.sort(array,new Comparator<Integer>(){    //从大到小排序                    @Override                    public int compare(Integer o1, Integer o2){                        return o2-o1;                    }                });                if(dfs(0,0,0)){                    System.out.println("yes");                }else{                    System.out.println("no");                }            }else{                System.out.println("no");            }        }    }    private static boolean dfs(int curLen, int num, int cur){        if(num == 3){            return true;        }        if(curLen == edge){            if(dfs(0,num+1,num+1)){                return true;            }else{                return false;            }        }else{             for(int i = cur; i < array.length; ++i){                if(!vis[i] && curLen+array[i] <= edge){                     vis[i] = true;                     if(dfs(curLen+array[i],num,i+1)){                         return true;                     }                     vis[i] = false;                }            }        }        return false;    }}