zoukankan      html  css  js  c++  java
  • POJ 2533 Longest Ordered Subsequence(LIS模版题)

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

    Description

    A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1a2, ..., aN) be any sequence (ai1ai2, ..., 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

    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
    ",*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
    ",mxlen);
        }
        return 0;
    }
  • 相关阅读:
    PIE SDK SFIM融合
    PIE SDK PCA融合
    c# 粘贴复制
    dev gridview 单元格值拖拽替换
    sql 行数据找出最大的及所有数据最大的
    mvc 登陆界面+后台代码
    mvc控制器返回操作结果封装
    Java 未来行情到底如何,来看看各界人士是怎么说的
    Java工程师修炼之路(校招总结)
    ​为什么我会选择走 Java 这条路?
  • 原文地址:https://www.cnblogs.com/Blackops/p/5853673.html
Copyright © 2011-2022 走看看