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Time limit : 3sec / Stack limit : 256MB / Memory limit : 256MB
Score : 700 points
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:
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.
The input is given from Standard Input in the following format:
N x1 x2 … xN L Q a1 b1 a2 b2 : aQ bQ
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.
9 1 3 6 13 15 18 19 29 31 10 4 1 8 7 3 6 7 8 5
4 2 1 2
For the 1-st query, he can travel from the 1-st hotel to the 8-th hotel in 4 days, as follows:
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|1 const 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; }
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原文地址：http://www.cnblogs.com/dyzll/p/5816213.html
