SquareTime Limit: 10000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9588Accepted Submission(s): 3127
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
开始就理解错题意了
o(╯□╰)o 为什么我总是理解错题意
题目的意思是所有的木棍能否组成一个正方形，而我认为所有木棍中的一部分是否可以构成一个正方形。。一直都是TLE，，我是枚举了所有的正方形可能的长度，然后进行深搜。。。
后来看了别人的代码才返现是自己理解错了。
即使题目意思明白了，我还是TLE一次。。原因是我重复搜索了。。
代码：#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int d[30] , m , sum = 0;bool visited[30] ;bool DFS(int len , int c ,int pos){if(c==4){return true ;}if(sum == len){if(DFS(0,c+1,0)){return true ;}}else{for(int i = pos ; i < m ; ++i){if(!visited[i]){if(len+d[i]>sum){return false;}visited[i] = true ;if(DFS(len+d[i],c,i+1)){return true ;}visited[i] = false ;}}}return false ;}int main(){int n ;scanf("%d",&n);while(n–){scanf("%d",&m);sum = 0 ;for(int i = 0 ; i < m ; ++i){scanf("%d",&d[i]) ;sum += d[i] ;}if(m<4 || sum%4!=0){puts("no") ;}else{sort(d,d+m) ;sum /= 4 ;if(sum<d[m-1]){puts("no") ;continue ;}memset(visited,false,sizeof(visited)) ;if( DFS(0,0,0) ){puts("yes") ;}else{puts("no") ;}}}return 0 ;}
，也有善意的提醒：何不去远方！昆明呀——赶一个花海;
