AtCoder Regular Contest 060 E - 高橋君とホテル / Tak and Hotels（倍增）

E - 高橋君とホテル / Tak and Hotels

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Time Limit: 3 sec / Memory Limit: 256 MB

Score : 700700 points

Problem Statement

NN hotels are located on a straight line. The coordinate of the ii-th hotel (1≤i≤N)(1≤i≤N) is xixi.

Tak the traveler has the following two personal principles:

• He never travels a distance of more than LL 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 QQ queries. The jj-th (1≤j≤Q)(1≤j≤Q) query is described by two distinct integers ajaj and bjbj. For each query, find the minimum number of days that Tak needs to travel from the ajaj-th hotel to the bjbj-th hotel following his principles. It is guaranteed that he can always travel from the ajaj-th hotel to the bjbj-th hotel, in any given input.

Constraints

• 2≤N≤1052≤N≤105
• 1≤L≤1091≤L≤109
• 1≤Q≤1051≤Q≤105
• 1≤xi<x2<...<xN≤1091≤xi<x2<...<xN≤109
• xi+1−xi≤Lxi+1−xi≤L
• 1≤aj,bj≤N1≤aj,bj≤N
• aj≠bjaj≠bj
• N,L,Q,xi,aj,bjN,L,Q,xi,aj,bj are integers.

Partial Score

• 200200 points will be awarded for passing the test set satisfying N≤103N≤103 and Q≤103Q≤103.

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Input

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

NN
x1x1 x2x2 ...... xNxN
LL
QQ
a1a1 b1b1
a2a2 b2b2
:
aQaQ bQbQ

Output

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

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Sample Input 1 Copy

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

Copy

4
2
1
2

For the 11-st query, he can travel from the 11-st hotel to the 88-th hotel in 44 days, as follows:

• Day 11: Travel from the 11-st hotel to the 22-nd hotel. The distance traveled is 22.
• Day 22: Travel from the 22-nd hotel to the 44-th hotel. The distance traveled is 1010.
• Day 33: Travel from the 44-th hotel to the 77-th hotel. The distance traveled is 66.
• Day 44: Travel from the 77-th hotel to the 88-th hotel. The distance traveled is 1010.

題意：

給你x軸上n(n<=1e5)個點的座標a[i](0<a[i]<a[i+1]<...<=1e9)，給你一個數l表示每次最多走長度l，且只能走到給你的點上。(a[i+1]-a[i]<=l)

Q次詢問，每次詢問兩個點x,y，問從x走到y最少需要多少步。

思路：

直接倍增，預處理每個點能走的最遠的點。

每次查詢直接跳就行。

代碼：

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(register int i=(a);i<=(b);i++)
#define dep(i,a,b) for(register int i=(a);i>=(b);i--)
using namespace std;
const int maxn=2e5+5;
//const double pi=acos(-1.0);
//const double eps=1e-9;
//const ll mo=1e9+7;
int n,m,k,l,q;
int a[maxn],c[maxn];
int f[maxn][25];
int ans,tmp,cnt;
int flag;
char s[maxn];
bool ok[maxn];
int main()
{
    int T,cas=1;
    //scanf("%d",&T);
    while(scanf("%d",&n)!=EOF)
    {
        rep(i,0,n) c[i]=1;
        int j=1;
        rep(i,1,n) scanf("%d",&a[i]);
        scanf("%d",&l);
        rep(i,0,20)
        for(int j=0;j<=n;j++)
        f[j][i]=inf;
        rep(i,1,n){
            while(a[i]>a[j]+l) {f[j][0]=i-1;j++;}
        }
        rep(i,1,20)
        {
            c[i]=c[i-1]<<1;
            for(int j=1;j<=n;j++)
            if(f[j][i-1]!=inf){
                //cout<<f[j][i-1]<<endl;
                f[j][i]=f[f[j][i-1]][i-1];
            }
        }
        ans=0;
        scanf("%d",&q);
        rep(t,1,q)
        {
            int x,y,ans=0;
            scanf("%d%d",&x,&y);
            if(x>y) swap(x,y);
            dep(i,20,0)
            {
                if(f[x][i]<y)
                {
                    x=f[x][i];
                    ans+=c[i];
                }
            }
            printf("%d\n",ans+1);
        }
    }
    return 0;
}