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): 86    Accepted Submission(s): 43

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

算出有多少个区间。满足最大值-最小值 < k

解法：

枚举每个位置x是区间的最左边位置。二分查找，最右的位置r，满足[x,r]的最大值-最小值恰好 < k

r-x+1就是解

用树状数组更新最大最小值，二分位置。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 100007
int tree1[maxn];
int tree2[maxn];
int lowbit(int i){
    return i&(-i);
}
void addmax(int p,int num){
    while(p < maxn){
        tree1[p] = max(tree1[p],num);
        p += lowbit(p);
    }
}

int querymax(int p){
    int ans = 0;
    while(p > 0){
        ans = max(ans,tree1[p]);
        p -= lowbit(p);
    }
    return ans;
}

void addmin(int p,int num){
    while(p < maxn){
        tree2[p] = min(tree2[p],num);
        p += lowbit(p);

    }
}
int querymin(int p){
    int ans = 1000000007;
    while(p > 0){
        ans = min(ans,tree2[p]);
        p -= lowbit(p);
    }
    return ans;

}

int num[maxn];
int main(){
    int t,n,k;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&k);
        for(int i = 1;i <= n; i++){
            scanf("%d",&num[i]);
        }
        memset(tree1,0,sizeof(tree1));
        memset(tree2,0x3f,sizeof(tree2));
        long long ans = 0;
        for(int i = n; i > 0 ; i--){
            addmax(i,num[i]);
            addmin(i,num[i]);
            int low = i,high = n;
            while(low <= high){
                int mid = (low+high)/2;
                int ma = querymax(mid);
                int mi = querymin(mid);
                if(ma - mi >= k)
                    high = mid-1;
                else
                    low = mid+1;
            }
            ans += low-i;
            //cout<<low<<endl;
        }
        cout<<ans<<endl;
    }
    return 0;
}