POJ 1887 Testing the CATCHER

 1 Source Code
 2 
 3 Problem: 1887  User: XXX
 4 Memory: 240K  Time: 63MS 
 5 Language: C++  Result: Accepted 
 6 
 7 Source Code 
 8 #include <stdio.h>
 9 int main()
10 {
11      int num[10000] ,total ,i ,j ,k ,Max ,max ,dp[10000] = {1};
12      for(k = 1 ; ;k++)
13      {
14             total = -1 ,Max = 1;
15             do{
16                  total++;
17                  scanf("%d",&num[total]);
18             }while(num[total] != -1);
19             if(total == 0) break;
20             for(i = 1 ;i < total ;i++)
21             {
22                   dp[i] = 1 ,max = 0;
23                   for(j = 0 ;j < i ;j++)
24                   {
25                         if(num[i] < num[j] && dp[j] > max)
26                             max = dp[j];
27                   }
28                   dp[i] = max + 1;
29                   if(dp[i] > Max) Max = dp[i];
30             }
31             printf("Test #%d:\n  maximum possible interceptions: %d\n\n" ,k ,Max);
32      }
33      return 0;
34 }

O(n2)的时间复杂度。。。。

O(nlnn)还不会。。。。

posted on 2012-02-26 11:30 小墨在努力！！阅读(150) 评论(0) 编辑收藏举报