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
题意：给定一个正整数序列，问有多少个连续子序列，其中最大值与最小值之差小于k？
题解：神奇的单调队列，用一个单调递增和一个单调递减队列来维护最小值与最大值，用两个前后指针，前指针指向当前区间左端点，后指针指向当前元素。如果当前加入新元素之后，max-min>=k，那么与新元素相差至少k的那个元素之前的所有元素都将失效，也就是出队。每一次操作答案都加上新元素向左可以延伸的最大距离。
例子：n=6，k=3，序列为5 3 5 2 3 4。q1为单调增队列，q2为单调减队列。
step1 :
q1 : 5
q2 : 5
l = 0, r = 0
ans = 1
step2 :
q1 : 3
q2 : 5 3
l = 0, r = 1
ans = 1 + 2
step3 :
q1 : 3 5
q2 : 5
l = 0, r = 2
ans = 1 + 2 + 3
step4 :
q1 : 2
q2 : 5 2
l = 0, r = 3
这时候发现q1中最小值与q2中最大值差值3>=k，那么开始出队。
q1 : 2
q2 : 2
l = 3, r = 3
ans = 1 + 2 + 3 + 1
step5 :
q1 : 2 3
q2 : 3
l = 3, r = 4
ans = 1 + 2 + 3 + 1 + 2
step6 :
q1 : 2 3 4
q2 : 4
l = 3, r = 5
ans = 1 + 2 + 3 + 1 + 2 + 3
所以最后答案为12
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
int a[100010];
deque<int> q1, q2;
int main()
{
int T;
cin >> T;
while (T--)
{
int n, k;
cin >> n >> k;
for (int i = 0; i < n; ++i)
scanf("%d", &a[i]);
q1.clear();
q2.clear();
int l = 0, r = 0;
long long ans = 0;
while (r < n)
{
while (!q1.empty() && q1.back() > a[r])
q1.pop_back();
q1.push_back(a[r]);
while (!q2.empty() && q2.back() < a[r])
q2.pop_back();
q2.push_back(a[r]);
while (!q1.empty() && !q2.empty() && q2.front() - q1.front() >= k)
{
if (q1.front() == a[l])
q1.pop_front();
if (q2.front() == a[l])
q2.pop_front();
l++;
}
ans += r - l + 1;
r++;
}
cout << ans << endl;
}
return 0;
}