Good subsequence( RMQ+二分)

Description

Give you a sequence of n numbers, and a number k you should find the max length of Good subsequence. Good subsequence is a continuous subsequence of the given sequence and its maximum value - minimum value<=k. For example n=5, k=2, the sequence ={5, 4, 2, 3, 1}. The answer is 3, the good subsequence are {4, 2, 3} or {2, 3, 1}.

Input

There are several test cases.
Each test case contains two line. the first line are two numbers indicates n and k (1<=n<=10,000, 1<=k<=1,000,000,000). The second line give the sequence of n numbers a[i] (1<=i<=n, 1<=a[i]<=1,000,000,000).
The input will finish with the end of file.

Output

For each the case, output one integer indicates the answer.

Sample Input

5 2
5 4 2 3 1
1 1
1

Sample Output

3
1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <sstream>
#include <iomanip>
using namespace std;
const int INF=0x4fffffff;
const int EXP=1e-6;
const int MS=10005;
const int MS2=100005;

int a[MS];
int n,k;

int minv[MS][15];
int maxv[MS][15];

void RMQ_init()
{
      for(int i=0;i<n;i++)
      {
            minv[i][0]=a[i];
            maxv[i][0]=a[i];
      }

      for(int j=1;(1<<j)<=n;j++)
      {
            for(int i=0;i+(1<<j)-1<n;i++)
            {
                  minv[i][j]=min(minv[i][j-1],minv[i+(1<<(j-1))][j-1]);
                  maxv[i][j]=max(maxv[i][j-1],maxv[i+(1<<(j-1))][j-1]);
            }
      }
}

int RMQ(int l,int r)
{
      int k=0;
      while(1<<(k+1)<=r-l+1)
            k++;
      return   max(maxv[l][k],maxv[r-(1<<k)+1][k])-min(minv[l][k],minv[r-(1<<k)+1][k]);
}

int main()
{

      while(scanf("%d%d",&n,&k)!=EOF)
      {
            for(int i=0;i<n;i++)
                  scanf("%d",&a[i]);
            RMQ_init();
            int ans=0;
            for(int i=0;i<n;i++)
            {
                  int l=i;
                  int r=n-1;
                  while(l<=r)
                  {
                        int mid=(l+r)/2;
                        int t=RMQ(i,mid);
                        if(t>k)
                              r=mid-1;
                        else
                              l=mid+1;
                  }
                  if(ans<l-i)
                        ans=l-i;
            }
            cout<<ans<<endl;
      }
      return 0;
}