Square（DFS）

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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

解题思路

n个木棒能否构成正方形且都要用上，按边来深搜，

当返回到最初的dfs中，for循环结束未能return 1;则返回值0到主函数中；

count是当前层的count，如当从第二条边回溯到第一条边时，此时count=0；

AC代码

#include<stdio.h>#include<string.h>int a[21];bool used[21];int n,m,sum;int dfs(int d, int w, int count){if(d == sum)                                          //d==sum时返回值1到上一层，进行return 1;再返回到上一层，{w=0;   d=0; count++;if(count == 3)return 1;}for(int i=w; i<m; i++)if(!used[i]){used[i] = 1;if((d+a[i] <= sum ) && dfs(d+a[i], i+1, count))                    return 1;used[i] = 0;}return 0;                                          //当前边不能构成，返回上一层}int main(){scanf("%d",&n);while(n--){scanf("%d",&m);sum=0;memset(used, 0, sizeof(used));for(int i=0; i<m; i++){scanf("%d",&a[i]);sum += a[i];}int flag=0;if(sum%4 == 0)                                  //构成正方形的先行条件{sum /= 4;  flag=dfs(0, 0, 0);}if(flag)printf("yes\n");elseprintf("no\n");}return 0;}

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