POJ 2533 Longest Ordered Subsequence（LIS模版题）-优快云博客

本文介绍了最长递增子序列(LIS)问题的两种解决方法：一种是O(N^2)的时间复杂度，另一种则是通过二分查找优化到O(NlogN)。这两种方法都用于寻找给定数列中最长递增子序列的长度。

Longest Ordered Subsequence

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 47465		Accepted: 21120

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

题目链接：POJ 2533

LIS模版题，N²和N*logN两种写法

N²代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1e3+10;
int arr[N],mx[N];
void init()
{
    CLR(arr,0);
    CLR(mx,0);
}
int main(void)
{
    int n,i,j,pre_len;
    while (~scanf("%d",&n))
    {
        init();
        for (i=1; i<=n; ++i)
            scanf("%d",&arr[i]);
        mx[1]=1;
        for (i=2; i<=n; ++i)
        {
            pre_len=0;
            for (j=1; j<i; ++j)
            {
                if(arr[j]<arr[i])//arr[i]可以接到arr[j]后面
                    if(mx[j]>pre_len)//接到一个具有最长LIS的后面。
                        pre_len=mx[j];
            }
            mx[i]=pre_len+1;
        }
        printf("%d\n",*max_element(mx+1,mx+1+n));
    }
    return 0;
}

NlogN代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1e3+10;
int arr[N],d[N];
void init()
{
    CLR(arr,0);
    CLR(d,0);
}
int main(void)
{
    int n,i,j,mxlen;
    while (~scanf("%d",&n))
    {
        init();
        for (i=1; i<=n; ++i)
            scanf("%d",&arr[i]);

        mxlen=1;
        d[mxlen]=arr[mxlen];

        for (i=2; i<=n; ++i)
        {
            if(d[mxlen]<arr[i])
                d[++mxlen]=arr[i];//最好情况一直往后增长
            else
            {
                int pos=lower_bound(d,d+mxlen,arr[i])-d;//用二分找到一个下界可放置位置
                d[pos]=arr[i];
            }
        }
        printf("%d\n",mxlen);
    }
    return 0;
}

转载于:https://www.cnblogs.com/Blackops/p/5853673.html