POJ2533(最长上升子序列)-优快云博客

本文介绍了一种使用动态规划解决最长递增子序列问题的经典算法。通过定义dp数组来记录以每个元素结尾的最长递增子序列长度，并最终找到整个序列中最长递增子序列的长度。

经典dp：设置数组dp[i]为如果取第i个数最长的个数，dp[0]为1，dp[i] 为max（dp[j]）(0<=j<i 且a[i]>a[j]),然后最后的答案为max（dp[i]）(0<=i<n)

题目描述：

Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 33577 Accepted: 14694
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

ac代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
#define N 1005
using namespace std;

int a[N], dp[N];
int main()
{
    int n;
    while(~scanf("%d", &n)){
        for(int i=0; i<n; i++)
            scanf("%d", &a[i]);

        memset(dp, 0, sizeof dp);
        dp[0] = 1;
        for(int i=0; i<n; i++){
            int tmax = 0;
            for(int j=0; j<i; j++){
                if(a[j] < a[i]){
                    if(dp[j] > tmax)
                        tmax = dp[j];
                }
            }
            dp[i] = tmax+1;
        }

        int ans = 0;
        for(int i = 0; i<n; i++){
            if(ans < dp[i])
                ans = dp[i];
        }
        printf("%d\n", ans);
    }
    return 0;
}