poj2553 Longest Ordered Subsequence

最新推荐文章于 2021-07-30 17:43:41 发布

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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，通过一个实例详细展示了算法的具体实现过程。

poj2553

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13831		Accepted: 5864

Description

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

思路：本题是求的最长上升子序列，这里用DP，状态转移方程为：

for 0->i to N-1

LIS[i]=1；

LIS[K]=MAX{LIS[i]:0<=i<k&&a[k]>a[i]}+1

代码：

#include<iostream>
using namespace std;
int a[1010];
int LIS[1010];
int main()
{
    int N,i,j,max;
scanf("%d",&N);
for(i=0;i<N;i++)
{
    scanf("%d",&a[i]);
}
LIS[0]=1;
for(i=0;i<N;i++)
{
    LIS[i]=1;
    for(j=0;j<=i;j++)
    {
        if(a[i]>a[j]&&LIS[j]+1>LIS[i])
            LIS[i]=LIS[j]+1;
    }
}
max=0;
for(i=0;i<N;i++)
{
if(LIS[i]>max)
max=LIS[i];
}
printf("%d/n",max);

return 0;
}