HDU 5289 Assignment（2015 Multi-University Training Contest 1）

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2976    Accepted Submission(s): 1385

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]  
    
    
    
    
     
   
   
   
   

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

题意：给T足数据，然后每组一个n和k，表示n个数，k表示最大允许的能力差，接下来n个数表示n个人的能力，求能力差在k之内的区间有几个。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define LL __int64
const int MIN=1e9+10;
const int MAX=0;
const int maxm=1e5+10;
int a[maxm];
int mina[maxm];
int maxa[maxm];
int n,k;
void Init()
{
    for(int i=0;i<=maxm;i++)
    {
        mina[i]=MIN;
        maxa[i]=MAX;
    }
}
int lowbit(int i)
{
    return i&(-i);
}
void add(int i,int v)
{
    while(i<=n)
    {
        mina[i]=min(mina[i],v);
        maxa[i]=max(maxa[i],v);
        i+=lowbit(i);
    }
}
int quertymin(int i)
{
    int Min=MIN;
    while(i>0)
    {
        Min=min(Min,mina[i]);
        i-=lowbit(i);
    }
    return Min;
}
int quertymax(int i)
{
    int Max=MAX;
    while(i>0)
    {
        Max=max(Max,maxa[i]);
        i-=lowbit(i);
    }
    return Max;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        Init();
        scanf("%d%d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        LL sum=0;
        for(int i=n;i>=1;i--)
        {
            add(i,a[i]);
            int low=i;
            int high=n;
            int p=0;
            while(low<=high)
            {
                int mid=(low+high)>>1;
                if(quertymax(mid)-quertymin(mid)<k)
                {
                    p=mid;
                    low=mid+1;
                }
                else
                {
                    high=mid-1;
                }
            }
            sum+=p-i+1;
        }
        printf("%I64d\n",sum);
    }
    return 0;
}