题解报告：poj 2533 Longest Ordered Subsequence（最长上升子序列LIS）

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
解题思路：典型dp：最长上升子序列问题。有两种解法，一种O(n^2)，另一种是O(nlogn)。相关详细的讲解：LIS总结。
AC代码一：朴素O(n^2)算法，数据小直接暴力。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=1005;
int n,res,dp[maxn],a[maxn];
int main(){
    while(~scanf("%d",&n)){
        memset(dp,0,sizeof(dp));res=0;
        for(int i=0;i<n;++i)scanf("%d",&a[i]),dp[i]=1;//每个自身都是一个长度为1的子序列
        for(int i=0;i<n;++i){
            for(int j=0;j<i;++j)
                if(a[j]<a[i])dp[i]=max(dp[i],dp[j]+1);//只包含i本身长度为1的子序列
            res=max(res,dp[i]);
        }
        printf("%d\n",res);
    }
    return 0;
}

AC代码二：进一步优化，采用二分法每次更新最小序列，最终最小序列的长度（其最终的序列不一定是正确的LIS，只是某个过程中有这个最长的序列，其长度就是最终<INF的元素个数）就是最长上升子序列长度。时间复杂度是O(nlogn)。dp[i]：长度为i+1的上升子序列中末尾元素的最小值（不存在的话就是INF），此处dp是针对相同长度下最小的末尾元素进行求解，角度转换十分巧妙。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=1005;
const int INF=0x3f3f3f3f;
int n,res,x,dp[maxn];
int main(){
    while(~scanf("%d",&n)){
        memset(dp,0x3f,sizeof(dp));
        for(int i=1;i<=n;++i){
            scanf("%d",&x);
            *lower_bound(dp,dp+n,x)=x;//更新最小序列
        }
        printf("%d\n",lower_bound(dp,dp+n,INF)-dp);
    }
    return 0;
}