POJ2533-Longest Ordered Subsequence

求解最长递增子序列

最新推荐文章于 2019-07-11 23:20:06 发布

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最新推荐文章于 2019-07-11 23:20:06 发布

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原文链接：http://www.cnblogs.com/lucio_yz/p/4719640.html

本文介绍了一种从后向前动态规划的方法来求解给定数列的最长递增子序列长度。通过实例演示了如何优化算法以避免冗余计算。

描述：

　　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.

　　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 file must contain a single integer - the length of the longest ordered subsequence of the given sequence.

代码：

　　求最长递增子序列的长度。从后向前动规，找当前数之后的大于当前数且动规值最大的值（即当前数能够连接成的最长序列的长度），动规值+1赋给当前值。充分体现了动态规划解决冗余的特点。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
using namespace std;
#define N 1005

int main(){
    int n,temp,dp[N],seq[N],count,flag,max;
    scanf("%d",&n);
    count=0;
    for( int i=0;i<n;i++ )
        scanf("%d",&seq[i]);
    max=0;
    for( int i=n-1;i>=0;i-- ){
        dp[i]=1;
        for( int j=i+1;j<n;j++ ){
            if( seq[j]>seq[i] && dp[i]<dp[j]+1 )
                dp[i]=dp[j]+1;
        }
        max=(max<dp[i])?dp[i]:max;
    }
    printf("%d\n",max);
    return 0;
}