Hdu 1518 Square, deep search, Pattern Pruning !!!, Hdu1518

Hdu 1518 Square, deep search, Pattern Pruning !!!, Hdu1518
SquareTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 9588 Accepted Submission (s): 3127


Problem DescriptionGiven a set of sticks of various lengths, is it possible to join them end-to-end to form a square?
 
InputThe 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.
 
OutputFor each case, output a line containing "yes" if 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

At the beginning, I understood the incorrect question meaning o (random □wrong) o why I always understood the incorrect question meaning that all wooden sticks can form a square, in my opinion, can a part of all wooden sticks form a square .. It's always TLE. I enumerated the possible lengths of all squares and then searched them in depth... After reading other people's code, I found myself wrong. Even if the question is clear, I will try it again .. The reason is that I have repeated searches .. Code:

#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 ;}