6463: Tak and Hotels II

6463: Tak and Hotels II

时间限制: 1 Sec  内存限制: 128 MB
提交: 95  解决: 29
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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.

来源

ABC044&ARC060

 题意：

给 n 个点 ， 一天最大的移动距离，要求每天结束时必须在点上，求从A点到B点的最小天数。

思路：

f[x][y] 表示第x个点第2^y天能到达的最远点，二分处理 f[i][0],  那么 f[i][j] = f[f[i][j-1]][j-1]（这里，因为j 表示 2的次方数， 所以第i个点 第 2^j 天能到达的最远点为  该点前 2^(j-1) 天能到达的最远点 经过 2^（j-1）天能到的最远点）。

记录一个从A 开始的 cur 值， 找到小于B的第一个值，改变 cur 值。

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n,a,b,p[maxn],l,Q;
int f[maxn][35];
int main(){
    scanf("%d",&n);
    for (int i=1; i<=n; i++) scanf("%d",&p[i]);
    scanf("%d",&l);
    for (int i=1;i<=n;i++){
        int idx=upper_bound(p+1,p+1+n,p[i]+l)-p-1;
        if(p[i]+l>=p[n]) f[i][0]=n;
        else 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];
    scanf("%d",&Q);
    while(Q--){
        scanf("%d%d",&a,&b);
        if(a>b) swap(a,b);
        int cur=a,ans=0;
        for (int i=30; i>=0; i--){
            if(f[cur][i]<b){
                ans+=(1<<i);
                cur = f[cur][i];
            }
        }
        printf("%d
",ans+1);
    }
    return 0;
}