2015 Multi-University Training Contest 1 Assignment

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 297    Accepted Submission(s): 152

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]

 

 

Source

2015 Multi-University Training Contest 1

 

解题：线段树+尺取法 或者 单调队列 然而我不会单调队列 也可以用st表

 

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 typedef long long LL;

 4 const int maxn = 100010;

 5 struct node {

 6     int lt,rt,minv,maxv;

 7 } tree[maxn<<2];

 8 int d[maxn];

 9 inline void pushup(int v) {

10     tree[v].minv = min(tree[v<<1].minv,tree[v<<1|1].minv);

11     tree[v].maxv = max(tree[v<<1].maxv,tree[v<<1|1].maxv);

12 }

13 void build(int lt,int rt,int v) {

14     tree[v].lt = lt;

15     tree[v].rt = rt;

16     if(lt == rt) {

17         tree[v].minv = tree[v].maxv = d[lt];

18         return;

19     }

20     int mid = (lt + rt)>>1;

21     build(lt,mid,v<<1);

22     build(mid+1,rt,v<<1|1);

23     pushup(v);

24 }

25 int query(int lt,int rt,int v,bool qmax) {

26     if(lt <= tree[v].lt && rt >= tree[v].rt)

27         return qmax?tree[v].maxv:tree[v].minv;

28     int ret = qmax?INT_MIN:INT_MAX;

29     if(lt <= tree[v<<1].rt)

30         ret = qmax?max(ret,query(lt,rt,v<<1,qmax)):min(ret,query(lt,rt,v<<1,qmax));

31     if(rt >= tree[v<<1|1].lt)

32         ret = qmax? max(ret,query(lt,rt,v<<1|1,qmax)):min(ret,query(lt,rt,v<<1|1,qmax));

33     return ret;

34 }

35 int main() {

36     int n,m,kase;

37     scanf("%d",&kase);

38     while(kase--) {

39         scanf("%d%d",&n,&m);

40         for(int i = 0; i < n; ++i)

41             scanf("%d",d+i);

42         build(0,n-1,1);

43         int low = 0,high = 0;

44         LL ret = 0;

45         while(low <= high && high < n) {

46             int minv = query(low,high,1,false);

47             int maxv = query(low,high,1,true);

48             if(maxv - minv < m) ret += high-low+1;

49             if(maxv - minv >= m) low++;

50             else high++;

51         }

52         cout<<ret<<endl;

53     }

54     return 0;

55 }

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