zoukankan      html  css  js  c++  java
  • HDU 5078 Revenge of LIS II(dp LIS)

    Problem Description
    In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
    ---Wikipedia

    Today, LIS takes revenge on you, again. You mission is not calculating the length of longest increasing subsequence, but the length of the second longest increasing subsequence.
    Two subsequence is different if and only they have different length, or have at least one different element index in the same place. And second longest increasing subsequence of sequence S indicates the second largest one while sorting all the increasing subsequences of S by its length.
     

    Input
    The first line contains a single integer T, indicating the number of test cases.

    Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.

    [Technical Specification]
    1. 1 <= T <= 100
    2. 2 <= N <= 1000
    3. 1 <= Ai <= 1 000 000 000
     

    Output
    For each test case, output the length of the second longest increasing subsequence.
     

    Sample Input
    3 2 1 1 4 1 2 3 4 5 1 1 2 2 2
     

    Sample Output
    1 3 2
    Hint
    For the first sequence, there are two increasing subsequence: [1], [1]. So the length of the second longest increasing subsequence is also 1, same with the length of LIS.
     

    Source
     

    Recommend

    參考链接:http://blog.csdn.net/acvay/article/details/40686171


    比赛时没有读懂题目開始做结果被hack了,后来一直想nlogn的方法,无果。以后应该会想出来,以后再贴那种方法代码


    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<vector>
    
    #define L(x) (x<<1)
    #define R(x) (x<<1|1)
    #define MID(x,y) ((x+y)>>1)
    
    #define eps 1e-8
    using namespace std;
    #define N 10005
    
    int a[N],dp[N],c[N];
    int n;
    
    int main()
    {
    	int i,j,t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%d",&n);
    		for(i=1;i<=n;i++)
    			scanf("%d",&a[i]);
    
            int ans=0;
            for(i=1;i<=n;i++)
    		{
    			dp[i]=1;
    			c[i]=1;
    			for(j=1;j<i;j++)
    			{
    			   if(a[i]<=a[j]) continue;
    
    			   if(dp[j]+1>dp[i])
    			   {
    			   	 dp[i]=dp[j]+1;
    			   	 c[i]=c[j];
    			   }
    			   else
    				if(dp[j]+1==dp[i])
    				 c[i]=2;
    			}
    			if(dp[i]>ans)
    				ans=dp[i];
    		}
    		j=0;
    		for(i=1;i<=n;i++)
    			if(dp[i]==ans)
    		{
    			j+=c[i];
    			if(j>1)
    				break;
    		}
    		if(j>1)
    			printf("%d
    ",ans);
    		else
    			printf("%d
    ",ans-1);
    	}
    	return 0;
    }
    




  • 相关阅读:
    C# 让程序自动以管理员身份运行
    [转]SAP算号器 license key Developer Access Key 完美解决方案
    【原创】项目性能优化全纪录(一) 存储过程优化
    treeview的遍历
    .NET求职笔试题目(续)
    SQL server 数据同步 Merge 的一个小bug
    use Stored procedure return a tabel(存储过程返回table)
    四种sql server 数据库分页的测试
    十五个世界最顶级的技术类博客网站
    层的拖动与隐藏
  • 原文地址:https://www.cnblogs.com/yfceshi/p/6931757.html
Copyright © 2011-2022 走看看