**Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 757    Accepted Submission(s): 282
**

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

这题想了很长时间，题意是求切去长度为l的连续子序列后，剩下的序列的最长上升子序列，切的起始位置随意定。

可以记录两个函数

f[i]:以a[i]为尾端的lis的最大长度。

g[i]:以a[i]为起始点的lis的最大长度。两者都包含a[i]。

这样对于每一个点，我们可以根据i, 找到 [0,i−L−1]之间的一个值，使得这个值小于a[i],且长度最大，可以用线段树来维护。然后用maxx=max(maxx,ans+g[i]),即前半段加上后半段就行了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100060
int f[maxn],g[maxn],pos[maxn],a[maxn],c[maxn],a1[maxn],h[maxn];
struct node{
    int l,r,maxnum;
}b[4*maxn];

void build(int l,int r,int i)
{
    int mid;
    b[i].l=l;b[i].r=r;b[i].maxnum=0;
    if(l==r)return;
    mid=(l+r)/2;
    build(l,mid,i*2);
    build(mid+1,r,i*2+1);
}

void update(int idx,int num,int i)
{
    int mid;
    if(b[i].l==idx && b[i].r==idx){
        b[i].maxnum=max(b[i].maxnum,num);return;
    }
    mid=(b[i].l+b[i].r)/2;
    if(idx<=mid)update(idx,num,i*2);
    else update(idx,num,i*2+1);
    b[i].maxnum=max(b[i*2].maxnum,b[i*2+1].maxnum);
}

int question(int l,int r,int i)
{
    int mid;
    if(l>r)return 0;
    if(b[i].l==l && b[i].r==r){
        return b[i].maxnum;
    }
    mid=(b[i].l+b[i].r)/2;
    if(r<=mid) return question(l,r,i*2);
    else if(l>mid)return question(l,r,i*2+1);
    else{
        int lv=question(l,mid,i*2);
        int rv=question(mid+1,r,i*2+1);
        return max(lv,rv);
    }

}

int main()
{
    int n,m,i,j,T,l,tot,len,maxx,ans,num1=0,k;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&l);
        tot=0;
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            pos[i]=a[i];
        }
        sort(pos+1,pos+1+n);
        tot=unique(pos+1,pos+1+n)-pos-1;
        for(i=1;i<=n;i++)a[i]=lower_bound(pos+1,pos+1+tot,a[i])-pos;
        f[1]=1;len=1;c[1]=a[1];
        for(i=2;i<=n;i++){
            if(a[i]>c[len]){
                len++;f[i]=len;
                c[len]=a[i];continue;
            }
            j=lower_bound(c+1,c+1+len,a[i])-c;
            c[j]=a[i];
            f[i]=j; //这里很重要，是j，不是len，因为求的是以a[i]为结尾的最长长度，不是前i个的最长长度
        }

        for(i=1;i<=n;i++){
            a1[i]=-a[n+1-i];
        }

        g[1]=1;len=1;c[1]=a1[1];
        for(i=2;i<=n;i++){
            if(a1[i]>c[len]){
                len++;g[i]=len;
                c[len]=a1[i];continue;
            }
            j=lower_bound(c+1,c+1+len,a1[i])-c;
            c[j]=a1[i];
            g[i]=j;
        }
        reverse(g+1,g+1+n);

        build(1,100000,1);
        maxx=0;
        for(i=l+1;i<=n;i++){
            ans=question(1,a[i]-1,1);
            update(a[i-l],f[i-l],1);
            maxx=max(maxx,ans+g[i]);
        }
        if(n-l>=1){
            maxx=max(maxx,f[n-l]);
        }
        num1++;
        printf("Case #%d: %d\n",num1,maxx);
    }
    return 0;
}
/*
2
10 2
1 2 3 9 8 7 4 5 6 7
*/