poj2553 Longest Ordered Subsequence

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poj2553

Longest Ordered Subsequence

Time Limit: 2000MS Memory Limit: 65536KTotal 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

71 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;
}