題目描述
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.
輸入
The input is given from Standard Input in the following format:
N
x1 x2 … xN
L
Q
a1 b1
a2 b2
:
aQ bQ
輸出
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.
樣例輸入

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

樣例輸出

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.

題意：給你直線上一些點的坐標，一天最大的移動距離，要求每天結束時必須在一個點上。1e5次詢問，每次詢問從第a個點到第b個點最少要幾天。

解題思路：

dp[x][y]表示第x個點2^y次方的天數能到的最遠點。暴力或二分預處理dp[x][0]，然后 dp[x][i] = dp[f[x][i-1]][i-1]；
查詢的時候, 找到從cur（當前點）出發的第一個小于b位置的dp[cur][j]，最后答案加上1（2^0）

//dp[x][y]表示第x個點2^ y次方的天數能到的最遠點。
$dp[j][i]=dp[dp[j][i?1]][i?1];$
// 含義以i等于1為例，意義是第j個點2天能到的最遠點。那么這不就是我一天最遠能走到的點，的一天能走到的最遠的點嘛
//其余的就可以依次類推

代碼：

int main() {while (cin >>n) {for (int i = 1; i <= n; ++i)cin >> a[i];cin >> L >> Q;for (int i = 1; i <= n; ++i) {//upper_bound()是找到第一個大于的位置，-a得到下標，再減1得到最后一個小于等于的點int id = upper_bound(a + 1, a + 1 + n, a[i] + L)-a-1;if (a[i] + L >= a[n]) //如果可以一次過那么在一天內就可以直接到n點了dp[i][0] = n;elsedp[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--) {cin >> x >> y;if (x > y) swap(x, y);ans = 0;//從大到小枚舉for (int i = 30; i >= 0; --i) {if (dp[x][i] < y) //從第一個點走的2^i的天數如果小于y，那么這個天數對于x點最優{ans += (1) << i;x = dp[x][i]; //如果還沒到終點以這個點來找} }cout << ans+1 << endl;} }return 0; }

lower_bound( )和upper_bound( )都是利用二分查找的方法在一個排好序的數組中進行查找的。

在從小到大的排序數組中，

lower_bound( begin,end,num)：從數組的begin位置到end-1位置二分查找第一個大于或等于num的數字，找到返回該數字的地址，不存在則返回end。通過返回的地址減去起始地址begin,得到找到數字在數組中的下標。

upper_bound( begin,end,num)：從數組的begin位置到end-1位置二分查找第一個大于num的數字，找到返回該數字的地址，不存在則返回end。通過返回的地址減去起始地址begin,得到找到數字在數組中的下標。

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