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高橋君とホテル / Tak and Hotels
高橋君とホテル / Tak and Hotels
Time limit : 3sec / Stack limit : 256MB / 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:
Nx1 x2 … xNLQa1 b1a2 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
91 3 6 13 15 18 19 29 311041 87 36 78 5
Sample Output 1
4212
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.
分析：对于每个点二分可以找到一天所能到达的最右和最左端点；
然后关键就是倍增，这样每个小问题就可以在log复杂度解决了；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000000#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=1e5+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m,k,t,l[30][maxn],r[30][maxn];ll a[maxn],p;void init(){ memset(l,-1,sizeof l); memset(r,-1,sizeof r); for(int i=0;i<=n-1;i++) { r[0][i]=lower_bound(a,a+n,a[i]+p)-a; if(r[0][i]==n||a[r[0][i]]>a[i]+p)r[0][i]--; l[0][i]=lower_bound(a,a+n,a[i]-p)-a; if(i==n-1)r[0][i]=-1; if(i==0)l[0][i]=-1; } for(int i=1;i<=29;i++) { for(int j=0;j<=n-1;j++) { if(r[i-1][j]<n-1)r[i][j]=r[i-1][r[i-1][j]]; if(l[i-1][j]>0)l[i][j]=l[i-1][l[i-1][j]]; } }}int main(){ int i,j; scanf("%d",&n); rep(i,0,n-1)scanf("%lld",&a[i]); scanf("%lld",&p); init(); int q; scanf("%d",&q); while(q--) { int b,c,ans=0; scanf("%d%d",&b,&c); b--,c--; if(b<c) { for(i=29;i>=0;i--) { if(r[i][b]!=-1&&r[i][b]<=c) { ans+=qpow(2,i); b=r[i][b]; if(b==c)break; } } if(i==-1)ans++; } else { for(i=29;i>=0;i--) { if(l[i][b]!=-1&&l[i][b]>=c) { ans+=qpow(2,i); b=l[i][b]; if(b==c)break; } } if(i==-1)ans++; } printf("%d\n",ans); } //system("Pause"); return 0;}
高橋君とホテル / Tak and Hotels