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.
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
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
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\n",ans+1); } return 0; }