zoukankan      html  css  js  c++  java
  • POJ 2533 Longest Ordered Subsequence DP

    Longest Ordered Subsequence
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 32192   Accepted: 14093

    Description

    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
    



    #include<iostream>
    #include<cstring>
    using namespace std;
    
    
    int main()
    {
    	int n;
    	while(cin>>n)
    	{
    		int a[1005],b[1005];
    		memset(b,0,sizeof(b));
    		int i,j;
    		for(i=0;i<n;i++)
    			cin>>a[i];
    		for(i=0;i<n;i++)
    		{
    			b[i]=1;
    			for(j=0;j<i;j++)
    				if(a[j]<a[i]&&b[i]<b[j]+1)
    					b[i]++;
    		}
    		int max=0;
    		for(i=0;i<n;i++)
    			if(max<b[i])    max=b[i];
    		cout<<max<<endl;
    	}
    	return 0;
    }


  • 相关阅读:
    《软件工程》第十六周学习进度
    个人总结
    构建之法阅读笔记06
    《软件工程》第十五周学习进度
    构建之法阅读笔记05
    《软件工程》第十四周学习进度
    买书问题
    第二阶段团队项目冲刺第十天
    第二阶段团队项目冲刺第九天
    第二阶段团队项目冲刺第八天
  • 原文地址:https://www.cnblogs.com/zhchoutai/p/6816135.html
Copyright © 2011-2022 走看看