zoukankan      html  css  js  c++  java
  • POJ 2533 Longest Ordered Subsequence(DP 最长上升子序列)

    Longest Ordered Subsequence
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 38980   Accepted: 17119

    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
    

    Source

    Northeastern Europe 2002, Far-Eastern Subregion



    n*n算法

    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    #include<queue>
    
    using namespace std;
    
    int n;
    int a[1001];
    int dp[1010];
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            int maxx = -1;
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
            }
            for(int i=0;i<n;i++)
            {
                dp[i] = 1;
                for(int j=0;j<i;j++)
                {
                    if(a[i]>a[j] && dp[j]+1>dp[i])
                    {
                        dp[i] = dp[j] + 1;
                    }
                }
                if(maxx < dp[i])
                {
                     maxx = dp[i];
                }
            }
            printf("%d
    ",maxx);
        }
        return 0;
    }




    n*logn算法:

    #include<iostream>
    #include<algorithm>
    #include<stdio.h>
    #include<string.h>
    #include<stdlib.h>
    
    #define inf 999999
    
    using namespace std;
    
    int n;
    int dp[1010];
    int a[1010];
    
    int res(int len,int num)
    {
        int l = 0,r = len;
        while(l!=r)
        {
            int mid = (l+r)>>1;
            if(dp[mid] == num)
            {
                return mid;
            }
            else if(dp[mid]<num)
            {
                l = mid + 1;
            }
            else if(dp[mid]>num)
            {
                r = mid;
            }
        }
        return l;
    }
    
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
            }
            int len = 1;
            dp[0] = -1;
            for(int i=1;i<=n;i++)
            {
                dp[i] = inf;
                int k = res(len,a[i]);
                if(k == len)
                {
                    len++;
                }
                dp[k] = a[i];
            }
            printf("%d
    ",len-1);
        }
        return 0;
    }


  • 相关阅读:
    OPENGL ES2.0如何不使用glActiveTexture而显示多个图片
    OpenGL帧缓存对象(FBO:Frame Buffer Object)
    EGLImage与纹理
    Android下Opengl ES实现单屏幕双眼显示
    comet4j开发指南
    tmp
    Ubuntu16.04下编译android6.0源码
    ubuntu下配置安装conky
    Qt编程之QImage类小结
    Linux学习,在线版
  • 原文地址:https://www.cnblogs.com/zsychanpin/p/6884550.html
Copyright © 2011-2022 走看看