最长上升子序列 (简单dp+二分)

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

O(n*n)：递推公式dp[i]=max(1,dp[j]+1)，j<i，ans=max(ans,dp[i]);

#include<bits/stdc++.h>
using namespace std;
int a[1005],dp[1005];
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            dp[i]=1;
            for(int j=1;j<i;j++)
                if(a[i]>a[j]) dp[i]=max(dp[i],dp[j]+1);
            ans=max(ans,dp[i]);
        }
        printf("%d\n",ans);
    }
}

O(nlogn): dp[i]表示为长度为i+1的最长上升子序列中末尾元素的最小值(初始值为INT_MAX)

#include<bits/stdc++.h>
using namespace std;
int dp[10005];
int main()
{
    int n,m;
    while(scanf("%d",&n)==1)
    {
        fill(dp,dp+n,INT_MAX);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&m);
            *lower_bound(dp,dp+n,m)=m;
        }
        printf("%d\n",lower_bound(dp,dp+n,INT_MAX)-dp);
    }
}