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  • POJ2533(KB12-N LIS)

      

    Longest Ordered Subsequence

    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 50827   Accepted: 22574

    Description

    A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., 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
     
     1 //2017-04-04
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <algorithm>
     6 
     7 using namespace std;
     8 
     9 const int N = 1005;
    10 int dp[N], a[N];//dp[i]表示前i个数字的最长上升子序列
    11 
    12 int main()
    13 {
    14     int n;
    15     while(cin>>n)
    16     {
    17         for(int i = 0; i < n; i++)
    18               cin>>a[i];
    19         int mx = 0;
    20         for(int i = 0; i < n; i++){
    21             dp[i] = 1;
    22             for(int j = 0; j < i; j++){
    23                 if(a[j] < a[i])
    24                       dp[i] = max(dp[i], dp[j]+1);
    25             }
    26             mx = max(mx, dp[i]);
    27         }
    28         cout<<mx<<endl;
    29     }
    30 
    31     return 0;
    32 }
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  • 原文地址:https://www.cnblogs.com/Penn000/p/6665310.html
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