POJ 2533 Longest Ordered Subsequence（单调递增子序列）

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35037		Accepted: 15372

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_iK), 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

4
大致题意：给出一个数列，求最大不下降子序列的长度。

思路：最优结构：dp[i]以第i个元素为子序列最后一个元素的LIS的长度。
状态转移方程：dp[1]=1;dp[i]=max(dp[j] | j<i && m[j] < m[i] ) +1 ;i>1

#include<iostream>
#include<cstdio>
using namespace std;
const int inf = 10001;
const int MAX = 1005;
int binary_search(int s[],int digit,int length)
{
    int left=0,right=length,mid;
    while(right!=left)
    {
        mid=(left+right)/2;
        if(digit==s[mid])
            return mid;
        else if(digit<s[mid])
            right=mid;
        else
            left=mid+1;
    }
    return left;
}
int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        int a[MAX];
        int s[MAX];
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        s[0]=-1;
        int len=1;
        for(i=1;i<=n;i++)
        {
            s[len]=inf;
            j=binary_search(s,a[i],len);
            if(j==len)
                len++;
            s[j]=a[i];
        }
        printf("%d
",len-1);
    }
    return 0;
}