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hdu 1518(dfs)

题目链接 :http://acm.hdu.edu.cn/showproblem.php?pid=1518

 

 

Square

 
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
 
 
 
 
题意 :给你n条边,让你用这些边组成正方形(不多不少仅n条)。
 
分析 :主要是超时问题,注意优化。
 
技术分享
 1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5  6 int a[22],vis[22]; 7 int n,m,length;               //length表示要组成的正方形的边长 8 int ans,flag; 9 10 void dfs(int cnt,int sum, int k)   //cnt记录边数,sum记录当前边长,k记录位置11 {12     if (cnt == 3)               //如果有三条边满足要求,那么第四条边一定满足要求13     {14         flag = 1;15         return ;16     }17     if (sum == length)        //找到满足要求的边,边数加一,初始化18     {19         cnt++;20         k = 0;21         sum = 0;22     }23     for (int i=k; i<m; i++)24     {25         if (!vis[i] && sum + a[i] <= length)26         {27             vis[i] = 1;28             dfs(cnt, sum+a[i], i+1);29             if (flag)         //优化时间,(当找到所有边之后就一直返回,不需要再把之后的代码运行一遍)30             {31                 return ;32             }33             vis[i] = 0;34         }35     }36 }37 38 int main ()39 {40     scanf ("%d",&n);41     while (n--)42     {43         ans = 0;44         scanf ("%d",&m);45         for (int i=0; i<m; i++)46         {47             scanf ("%d",&a[i]);48             ans += a[i];49         }50         if (ans % 4)              //如果所有边长和不是4的倍数,怎样都不能组成正方形51         {52             printf ("no\n");53             continue ;54         }55         memset(vis, 0, sizeof(vis));56         flag = 0;57         length = ans / 4;     //记录正方形边长58         dfs(0, 0, 0);59         if (flag)60             printf ("yes\n");61         else62             printf ("no\n");63     }64     return 0;65 }
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hdu 1518(dfs)