POJ2533——动规解决最长上升子序列

很典型的动态规划入门例题。

原序列a，dp[i]表示以a[i]结尾的最长上升子序列的长度。

初始状态：dp[i] = 1，表示将a[i]插进结果序列的末端。

两个循环：外层为 i：1 ～ N ；内层为 j：1 ～ i。

对应每个i和j，如果有a[i]>a[j]，那么dp[i]有两种选择：

1. 将dp[j]对应的序列插到a[i]前面；（此时dp[j]已是一个求好的最长上升子序列）

2. 选择不插进去。

于是有这样的状态转移方程：

dp[i] = max( dp[i] , dp[j] + 1 )，即max{ 不插进去，插进去（dp[j]+1）}

最后，要求出所有dp[i]的最大值，这个最大值才是所求最长上升子序列的长度。

题目：

A numeric sequence of a_i is ordered if a₁ < a₂ < ... < a_N. Let the subsequence of the given numeric sequence ( a₁, a₂, ..., a_N) be any sequence ( a_i₁, a_i₂, ..., a_{i_K}), where 1 <= i₁ < i₂ < ... < i_K <= 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

代码：

//
//  main.cpp
//  POJ2533
//
//  Created by jinyu on 2018/7/15.
//  Copyright © 2018年 jinyu. All rights reserved.
//

#include <iostream>
#include <algorithm>
using namespace std;
int a[1000+77];
int dp[1000+77];
int main(){
    
    int N,res;
    while(cin>>N){
        memset(dp, 0, sizeof(dp));
        res = 0;
        for(int i = 0;i<N;i++){
            cin>>a[i];
        }
        for(int i = 0;i<N;i++){
            dp[i] = 1;
            for(int j = 0;j<i;j++){
                if(a[i] > a[j]){
                    dp[i] = max(dp[i] , dp[j] + 1);
                }
            }
        }
        for(int i = 0;i<N;i++){
            if(dp[i] > res){
                res = dp[i];
            }
        }
        cout<<res<<endl;
    }
    return 0;
}

POJ2533——动规解决最长上升子序列

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