AtCoder Regular Contest 060 - 高橋君とホテル / Tak and Hotels - 倍增+二分

题目链接： https://arc060.contest.atcoder.jp/tasks/arc060_c

题目：

E - 高橋君とホテル / Tak and Hotels

Time limit : 3sec / Memory limit : 256MB

Score : 700 points

Problem Statement

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.

Input

The input is given from Standard Input in the following format:

N
x1 x2 … xN
L
Q
a1 b1
a2 b2
:
aQ bQ

Output

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.

Sample Input 1

Copy

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

Sample Output 1

Copy

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n个点然后给你每个点的坐标(抽象成一条直线) 给你每天走的最远的距离L 和 Q次询问输出每次询问中L点到R点的最少天数，如果暴力的话就是Q*N*lgN 每次问询 * 最短时间 TLE

然后看到别人写的是倍增+二分就是提前预处理好2^x天数可达的最远距离

可以用dp求解 dp[i][j] = dp[dp[i][j-1]][j-1] 就是第i个点第2^j天等于第i个点第2^(j-1)天到达的点第2^(j-1)天的位置

2^j = 2^(j-1) * 2 = 2^(j-1) + 2^(j-1);

先预处理2^0天的时间就是第一天可达的最远距离所以最后结果需要+1

最后求结果的时候从后往前找到L可达的最远小于等于R的某点~~~~~~然后不断的求X-R的最短时间求和即可

最后这样做的时间复杂度是(Q+N)*30 3e6

#include<bits/stdc++.h>
using namespace std;
 
int a[100005];
int dp[100005][50];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    int L,Q;
    scanf("%d%d",&L,&Q);
    for(int i = 1;i <= n;i++)
    {
        int id = upper_bound(a+1,a+1+n,a[i]+L)-a-1;
        if(a[i] + L >= a[n])
            dp[i][0] = n;
        else
            dp[i][0] = id;
    }
     
    for(int i = 1;i <= 30;i++)
    {
        for(int j = 1;j <= n;j++)
            dp[j][i] = dp[dp[j][i-1]][i-1];
    }
    while(Q--)
    {
        int L,R;
        scanf("%d%d",&L,&R);
        if(R<L)
            swap(L,R);  
        long long ans=0;
        for(int i = 30;i >= 0;i--)
        {
            if(dp[L][i] < R)
            {
                ans=ans+(1ll<<i);
                L = dp[L][i];
            }
        }
        printf("%lld\n",ans+1);
    }
    return 0;
}