HDU 5289 Assignment rmq

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#数据结构与算法 #php

本文探讨了如何解决分组问题，即在给定一组数值的情况下，找到满足特定条件的连续子数组数量。通过引入RMQ数据结构，实现了高效的线性时间复杂度解决方案。详细阐述了问题背景、输入输出规范、核心算法思想以及具体的代码实现。

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Assignment

题目连接：

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

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],…,an,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

题意

给你n个数，然后连续的，且这个区间内最大值减去最小值的差小于k的话，就可以算作一队。

问你一共有多少种分队的方法。

题解：

暴力枚举左端点位置，然后二分枚举右端点，用一个rmq维护一下就好了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
struct RMQ{
    const static int RMQ_size = maxn;
    int n;
    int ArrayMax[RMQ_size][21];
    int ArrayMin[RMQ_size][21];

    void build_rmq(){
        for(int j = 1 ; (1<<j) <= n ; ++ j)
            for(int i = 0 ; i + (1<<j) - 1 < n ; ++ i){
                ArrayMax[i][j]=max(ArrayMax[i][j-1],ArrayMax[i+(1<<(j-1))][j-1]);
                ArrayMin[i][j]=min(ArrayMin[i][j-1],ArrayMin[i+(1<<(j-1))][j-1]);
            }
    }

    int QueryMax(int L,int R){
        int k = 0;
        while( (1<<(k+1)) <= R-L+1) k ++ ;
        return max(ArrayMax[L][k],ArrayMax[R-(1<<k)+1][k]);
    }

    int QueryMin(int L,int R){
        int k = 0;
        while( (1<<(k+1)) <= R-L+1) k ++ ;
        return min(ArrayMin[L][k],ArrayMin[R-(1<<k)+1][k]);
    }


    void init(int * a,int sz){
        n = sz ;
        for(int i = 0 ; i < n ; ++ i) ArrayMax[i][0] = ArrayMin[i][0] = a[i];
        build_rmq();
    }

}solver;
int a[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;scanf("%d%d",&n,&k);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        solver.init(a,n);
        long long ans = 0;
        for(int i=0;i<n;i++)
        {
            int l = i,r = n-1,Ans=i;
            while(l<=r)
            {
                int mid = (l+r)/2;
                if(solver.QueryMax(i,mid)-solver.QueryMin(i,mid)<k)l=mid+1,Ans=mid;
                else r=mid-1;
            }
            ans += (Ans-i+1);
        }
        cout<<ans<<endl;
    }
    return 0;
}