hdu5489 ||2015合肥网络赛1006 dp+离散化树状数组优化

http://acm.hdu.edu.cn/showproblem.php?pid=5489

Removed Interval

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 656    Accepted Submission(s): 245

Problem Description

Given a sequence of numbers  A=a1,a2,…,aN , a subsequence  b1,b2,…,bk  of  A  is referred as increasing if  b1<b2<…<bk . LY has just learned how to find the longest increasing subsequence (LIS).
Now that he has to select  L  consecutive numbers and remove them from  A  for some mysterious reasons. He can choose arbitrary starting position of the selected interval so that the length of the LIS of the remaining numbers is maximized. Can you help him with this problem?

 

Input

The first line of input contains a number  T  indicating the number of test cases ( T≤100 ).
For each test case, the first line consists of two numbers  N  and  L  as described above ( 1≤N≤100000,0≤L≤N ). The second line consists of  N  integers indicating the sequence. The absolute value of the numbers is no greater than  109 .
The sum of N over all test cases will not exceed 500000.

 

Output

For each test case, output a single line consisting of “Case #X: Y”.  X  is the test case number starting from 1.  Y  is the maximum length of LIS after removing the interval.

 

Sample Input

   
   
   
   

    
    
    
    2
5 2
1 2 3 4 5
5 3
5 4 3 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 3
Case #2: 1
   
   
   
   
   
   
   
   


   
   
   
   

/**
hdu5489 ||2015合肥网络赛1006  dp+树状数组优化
题目大意：给定一个序列，求去掉一个连续长度为L的数后的最长上升子序列的长度
解题思路：用dp[i][0]表示前i个数不去L时的最长子序列，dp[i][1]表示前i个数去L时的最长子序列。
          转移方程为：dp[i][0]:它可以由0到i-1中满足权值<=x_i的dp0_j转移到.
                       dp[i][1]:它可以由0到i-1中满足权值<=x_i的dp1_j以及0到i-L-1中满足权值<=x_i的dp0_j转移到
          由于数据范围是10^5，需要离散化一下数字，用树状数组维护。处理很巧妙，详见代码
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=100100;
int a[maxn],num[maxn],n,m,L;
int dp[maxn][2],A[maxn],B[maxn];
bool cmp(int x,int y)
{
    return a[x]<a[y];
}
void add(int *A,int k1,int k2)
{
    for(; k1<=m; k1+=(k1&(-k1)))A[k1]=max(A[k1],k2);
}
int find(int *A,int k1)
{
    int ans=-1e9;
    for(; k1; k1-=(k1&(-k1)))ans=max(ans,A[k1]);
    return ans;
}
int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&L);
        for(int i=1; i<=n; i++)scanf("%d",&a[i]);
        a[0]=-2e9;
        for(int i=0; i<=n; i++)num[i]=i;
        sort(num,num+n+1,cmp);
        m=1;
        int pre=a[num[0]];
        a[num[0]]=m;
        for(int i=1; i<=n; i++)
        {
            if(a[num[i]]!=pre)m++,pre=a[num[i]];
            a[num[i]]=m;
        }
//        for(int i=0;i<=n;i++) printf("%d ",a[i]);
//        printf("\n");
        for(int i=1; i<=m; i++)A[i]=0,B[i]=-1e9;
        if(L==0)for(int i=1; i<=m; i++)B[i]=0;
        for(int i=1; i<=n; i++)
        {
            dp[i][0]=find(A,a[i]-1)+1;
            dp[i][1]=find(B,a[i]-1)+1;
            add(A,a[i],dp[i][0]);
            add(B,a[i],dp[i][1]);
            if(i>=L) add(B,a[i-L],max(dp[i-L][0],dp[i-L][1]));
        }
        int ans=max(0,dp[n-L][0]);
        for(int i=1; i<=n; i++)ans=max(ans,dp[i][1]);
        printf("Case #%d: %d\n",++tt,ans);
    }
    return 0;
}