POJ - 2533----Longest Ordered Subsequence

最新推荐文章于 2021-07-25 22:06:01 发布

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最新推荐文章于 2021-07-25 22:06:01 发布

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分类专栏：动态规划文章标签： dp

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本文介绍了一种使用动态规划解决最长上升子序列问题的方法，并提供了一个示例程序，该程序能够找出给定序列中最长上升子序列的长度。

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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

题目大意：求一个序列中的最长上升子序列的长度

思路：
dp[i]为以序列中第i个元素结尾的最长上升子序列的长度
则状态转移方程为
if(A[i]>A[j]) dp[i]=max(dp[i],dp[j] + 1)

#include<iostream>  
#include<cstdio>  
#include<cstring>  
#include<algorithm>
using namespace std;
const int MAX=1001;

int A[MAX],dp[MAX];
int N,ans;

int main(void){
    while(cin>>N){
        for (int i=0 ; i<N ; i++){
            cin>>A[i]; 
            dp[i]=1;
        }
        ans=1;
        for(int i=0 ; i<N ; i++){
            for(int j=0 ; j<i ; j++){
                if(A[j] < A[i]){
                    dp[i]=max(dp[i],dp[j]+1);
                }
            }
            ans=max(ans,dp[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}