hdu 1518 Square 深搜,,,,花样剪枝啊!!!
Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9588 Accepted 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
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
開始就理解错题意了
o(╯□╰)o 为什么我总是理解错题意
题目的意思是全部的木棍能否组成一个正方形,而我觉得全部木棍中的一部分能否够构成一个正方形。。
一直都是TLE,,我是枚举了全部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。
即使题目意思明确了。我还是TLE一次。。原因是我反复搜索了。。
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 ; }
posted on 2017-06-12 16:27 cynchanpin 阅读(171) 评论(0) 编辑 收藏 举报