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。

 

 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         else

62             printf ("no\n");

63     }

64     return 0;

65 }

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