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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ 2533 Longest Ordered Subsequence

POJ 2533 Longest Ordered Subsequence

編輯:C++入門知識

Longest Ordered Subsequence Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 25229 Accepted: 10957 Description   A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).   Your program, when given the numeric sequence, must find the length of its longest ordered subsequence. Input   The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000 Output   Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence. Sample Input   7 1 7 3 5 9 4 8 Sample Output   4 Source   Northeastern Europe 2002, Far-Eastern Subregion 最長上升子序列 [cpp]   #include <stdio.h>   #include <math.h>   #include <string.h>   int a[1100],n;   int main()   {       int find();       int i,j,m,s,t,max;       scanf("%d",&n);       for(i=0;i<=n-1;i++)       {           scanf("%d",&a[i]);       }       max=find();       printf("%d\n",max);       return 0;   }   int find()   {       int i,j,dp[1100],max;       memset(dp,0,sizeof(dp));       dp[0]=1;       for(i=1;i<=n-1;i++)       {           max=0;           for(j=0;j<=i-1;j++)           {               if(a[i]>a[j])               {                   if(dp[j]>max)                   {                       max=dp[j];                   }               }           }           dp[i]=max+1;       }       for(i=0,max=0;i<=n-1;i++)       {           if(dp[i]>max)           {               max=dp[i];           }       }       return max;   }  

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