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

Assignment

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

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

Sample Output

5
28

Hint

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

题意：

问有多少区间段，最大小值差<k。

做法：

枚举右端点，很明显 区间越大，最大小值差越大，所以有线性关系。所以可以二分。找到差值小于k的点，这个点到右端点之间所有点都可以做为左端点。

线段树和树状数组都可能超时，离线最大小值计算最稳的就是RMQ了。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN = 100100;
int n,query;
int A[MAXN];
int FMin[MAXN][20],FMax[MAXN][20];

void Init(){
	int i,j;
	for(i=1;i<=n;i++)
		FMin[i][0]=FMax[i][0]=A[i];
	for(i=1;(1<<i)<=n;i++){   //按区间长度递增顺序递推
		for(j=1;j+(1<<i)-1<=n;j++){   //区间起点
			FMin[j][i]=min(FMin[j][i-1],FMin[j+(1<<(i-1))][i-1]);
			FMax[j][i]=max(FMax[j][i-1],FMax[j+(1<<(i-1))][i-1]);
		}
	}
}

int Query(int l,int r){
	int k=(int)(log(double(r-l+1))/log((double)2));
	return max(FMax[l][k],FMax[r-(1<<k)+1][k]);
}
int Query2(int l,int r){
	int k=(int)(log(double(r-l+1))/log((double)2));
	return min(FMin[l][k],FMin[r-(1<<k)+1][k]);
}
int main(){
	int a,b;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int k;
		scanf("%d%d",&n,&k);
		for(int i=1;i<=n;i++) scanf("%d",&A[i]);
		Init();
		int lll=1;
		__int64 ans=0;
		for(int i=1;i<=n;i++)
		{
			int l=lll;
			int r=i;
			while(l<=r)
			{
				int mid=(l+r)/2;
				int low=Query2(mid,i);
				int hig=Query(mid,i);
				int tt=hig-low;

				if(tt>=k)
					l=mid+1;
				else if(tt<k)
					r=mid-1;
			}

			lll=l;
			ans+=i-l+1;
		}

		printf("%I64d\n",ans);
	}
	return 0;
}

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