HDU 5289 Assignment RMQ

Assignment

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

 

题意：

给你一个长度为n，的序列和一个k，让你找出有多少个[l,r]区间满足： 该区间的最大最小值之差小于k

题解：

我们固定左端点l，二分其右端点，用RMQ维护最大最小值就好了

 

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e5+20, M = 30005, mod = 1000000007, inf = 0x3f3f3f3f;
typedef long long ll;
//不同为1，相同为0
int dp[N][22],f[N][22],a[N],n,T,k;
void RMQ() {
    for(int i = 1; i <= n; i++) dp[i][0] = f[i][0] = a[i];
    for(int i = 1; (1<<i) <= n; i++) {
        for(int j = 1; (j + (1<<i) - 1) <= n; j++) {
            dp[j][i] = max(dp[j][i-1],dp[j + (1<<(i-1))][i-1]);
            f[j][i] = min(f[j][i-1],f[j + (1<<(i-1))][i-1]);
        }
    }
}
int query(int l,int r,int p) {
    if(l == r) return a[l];
    int k = (int) ((log((double)(r - l + 1))) / log(2.0));
    if(p) return max(dp[l][k] , dp[r - (1<<k) + 1][k]);
    else return min(f[l][k] , f[r - (1<<k) + 1][k]);
}
int main() {
  scanf("%d",&T);
  while(T--) {

    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    RMQ();
    ll ans = 0 ;
    for(int l=1;l<=n;l++) {
        int L = l, R = n,A=l;
        while(L<=R) {
            int mid = (L+R)>>1;
            if(query(l,mid,1)-query(l,mid,0)<k) L = mid+1,A=mid;
            else  R = mid-1;
        }
        ll sum = A-l+1;
        ans+=(sum);
    }
   printf("%I64d
",ans);
  }
  return 0;
}