HDU-1518 Square(DFS)

Square

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 12

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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

Source

University of Waterloo Local Contest 2002.09.21

此题需要优化时间，避免超时。。优化时间技巧可以学习。。。。。。。。。

 #include <stdio.h>
 #include<string.h>
 int a[];
 int vist[];
 int sum;
 int l;
 int n;
 int flag;
 void Dfs(int t, int len, int index)
 {

     if (t == )
     {
         flag = ;
         return ;
     }

     if (len == l)
     {
         Dfs(t + , , );
         if (flag)//优化时间
         {
             return ;
         }
     }

     for (int i = index; i < n; i++)//从index开始优化时间
     {
         if (vist[i]== && a[i] + len <= l)
         {
             vist[i] = ;
             Dfs(t, a[i] + len, i + );
             if (flag)//优化时间
             {
                 return;
             }
             vist[i] = ;
         }
     }
 }

 int main()
 {
     int i,t;
     scanf("%d", &t);
     while (t--)
     {

         sum = ;
         scanf("%d", &n);
         for (int i = ; i < n; i++)
         {
             scanf("%d", &a[i]);
             sum += a[i];
         }

         if (sum %  != )//简答的优化
         {
             puts("no");
             continue;
         }

         l = sum / ;

         for (i = ; i < n; i++)//有比边长大的边就不行
         {
             if (a[i] > l)
             {
                 break;
             }
         }
         if (i != n)
         {
             puts("no");
             continue;
         }
        memset(vist, , sizeof(vist));
         flag = ;
         Dfs(, , );
         if (flag)
         {
             puts("yes");
         }
         else
         {
             puts("no");
         }
     }
     return ;
 }