题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5289


题面：

Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 672    Accepted Submission(s): 335


Problem Description
Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. Tom want to know the number of groups like this.
 

Input
In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1<=n<=100000, 0<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,a[n](0<=a[i]<=10^9),indicate the i-th staff’s ability.
 

Output
For each test，output the number of groups.
 

Sample Input
2
4 2
3 1 2 4
10 5
0 3 4 5 2 1 6 7 8 9




 







5
28



Hint
First Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3] 

 





 


Author


FZUACM


 


Source








解题：


    比赛的时候，怎么想都想不正确。想去找近期的不合法的点，复杂度太高。


看了题解才知道是用ST算法的。先前不知道，这是一篇非常不错的ST算法的介绍。


javascript:void(0)


    枚举左边端点，二分右端点。用ST算法推断该区间是否合法，直至右端点到极限（即二分的左右边界相遇或交叉）。





代码：




#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <vector>
#include <cmath>
#include <algorithm>
#define mod 1000000007
using namespace std;
int t,n,k;
int a[100100],minn[100010][20],maxn[100010][20],mid;
long long ans;
void Rmq_Init()
{  
    int m=19;  
    for(int i=1;i<=n;i++) 
        maxn[i][0]=minn[i][0]=a[i];  
    for(int i=1;i<=m;i++)  
        for(int j=n;j;j--)
        {  
            maxn[j][i]=maxn[j][i-1];  
            minn[j][i]=minn[j][i-1];  
            if(j+(1<<(i-1))<=n)
            {  
                maxn[j][i]=max(maxn[j][i],maxn[j+(1<<(i-1))][i-1]);  
                minn[j][i]=min(minn[j][i],minn[j+(1<<(i-1))][i-1]);  
            }  
        }  
}  
int Query_dif(int l,int r)
{  
    int m=floor(log((double)(r-l+1))/log(2.0));  
    int Max=max(maxn[l][m],maxn[r-(1<<m)+1][m]);  
    int Min=min(minn[l][m],minn[r-(1<<m)+1][m]);  
    return Max-Min;  
}  
int solve(int l)
{
    int le,ri;
    le=l;
    ri=n;
        while(le<=ri)
        {
            mid=(le+ri)/2;
            if(Query_dif(l,mid)>=k)
            {
                ri=mid-1;
            }
            else
            {
                le=mid+1;
            }
        }
        /*if(Query_dif(l,mid)>=k)
          return mid-l;
        else 
          return mid-l+1;*/
        return ri-l+1;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
      ans=0;
      scanf("%d%d",&n,&k);
      for(int i=1;i<=n;i++)
      {
          scanf("%d",&a[i]);
      }
      Rmq_Init();
      for(int i=1;i<=n;i++)
      {
        ans=ans+solve(i);
        //cout<<i<<" "<<Query_dif(i,n)<<endl;
        //cout<<ans<<endl;
      }
      printf("%lld\n",ans);
    }
    return 0;
}