【倍增】Tak and Hotels II @ABC044&ARC060/upcexam6463

6463: Tak and Hotels II

时间限制: 1 Sec   内存限制: 128 MB

题目描述

N hotels are located on a straight line. The coordinate of the i-th hotel (1≤i≤N) is xi.
Tak the traveler has the following two personal principles:
He never travels a distance of more than L in a single day.
He never sleeps in the open. That is, he must stay at a hotel at the end of a day.
You are given Q queries. The j-th (1≤j≤Q) query is described by two distinct integers aj and bj. For each query, find the minimum number of days that Tak needs to travel from the aj-th hotel to the bj-th hotel following his principles. It is guaranteed that he can always travel from the aj-th hotel to the bj-th hotel, in any given input.

Constraints
2≤N≤105
1≤L≤109
1≤Q≤105
1≤xi<x2<…<xN≤109
xi+1−xi≤L
1≤aj,bj≤N
aj≠bj
N,L,Q,xi,aj,bj are integers.
Partial Score
200 points will be awarded for passing the test set satisfying N≤103 and Q≤103.

输入

The input is given from Standard Input in the following format:
N
x1 x2 … xN
L
Q
a1 b1
a2 b2
:
aQ bQ

输出

Print Q lines. The j-th line (1≤j≤Q) should contain the minimum number of days that Tak needs to travel from the aj-th hotel to the bj-th hotel.

样例输入

9
1 3 6 13 15 18 19 29 31
10
4
1 8
7 3
6 7
8 5

样例输出

4
2
1
2

提示

For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:
Day 1: Travel from the 1-st hotel to the 2-nd hotel. The distance traveled is 2.
Day 2: Travel from the 2-nd hotel to the 4-th hotel. The distance traveled is 10.
Day 3: Travel from the 4-th hotel to the 7-th hotel. The distance traveled is 6.
Day 4: Travel from the 7-th hotel to the 8-th hotel. The distance traveled is 10.

题意：给你直线上一些点的坐标，一天最大的移动距离，要求每天结束时必须在一个点上。1e5次询问，每次询问从第a个点到第b个点最少要几天。 solution f[x][y]表示第x个点2^y次方的天数能到的最远点。暴力或二分预处理f[x][0]，然后 f[x][i] = f[f[x][i-1]][i-1]； 查询的时候, 找到从cur（当前点）出发的第一个小于b位置的f[cur][j]，最后答案加上1（2^0） code

#define IN_LB() freopen("C:\\Users\\acm2018\\Desktop\\in.txt","r",stdin)
#define OUT_LB() freopen("C:\\Users\\acm2018\\Desktop\\out.txt","w",stdout)
#define IN_PC() freopen("C:\\Users\\hz\\Desktop\\in.txt","r",stdin)
#include <bits/stdc++.h>
using namespace std;
  
typedef long long ll;
const int maxn = 100005;
int ind[maxn],lnh,qry;
const int INF = 1<<30;
int f[maxn][35];
  
int main() {
//    IN_LB();
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++) {
        scanf("%d",ind+i);
    }
    scanf("%d%d",&lnh,&qry);
    for(int i=1; i<=n; i++) {
        int idx = (int)(upper_bound(ind+1,ind+n+1, ind[i]+lnh)-ind-1);
        f[i][0] = idx;
    }
    for(int j=1; j<=30; j++) {
        for(int i=1; i<=n; i++) {
            f[i][j] = f[f[i][j-1]][j-1];
        }
    }
  
    for(int q=0; q<qry; q++) {
        int a,b;
        scanf("%d%d",&a,&b);
        if(a>b)
            swap(a,b);
        int ans = 0,cur = a;
        for(int i=30;i>=0;i--){
            if(f[cur][i]<b){
                ans+=(1<<(i));
                cur = f[cur][i];
            }
        }
        printf("%d\n",ans+1);
    }
    return 0;
}