In the year of 30XX30XX participants of some world programming championship live in a single large hotel. The hotel has nn floors. Each floor has mm sections with a single corridor connecting all of them. The sections are enumerated from 11 to mm along the corridor, and all sections with equal numbers on different floors are located exactly one above the other. Thus, the hotel can be represented as a rectangle of height nn and width mm. We can denote sections with pairs of integers (i,j)(i,j), where ii is the floor, and jj is the section number on the floor.
The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections (1,x)(1,x), (2,x)(2,x), ……, (n,x)(n,x) for some xx between 11 and mm. All sections not occupied with stairs or elevators contain guest rooms. It takes one time unit to move between neighboring sections on the same floor or to move one floor up or down using stairs. It takes one time unit to move up to vvfloors in any direction using an elevator. You can assume you don‘t have to wait for an elevator, and the time needed to enter or exit an elevator is negligible.
You are to process qq queries. Each query is a question "what is the minimum time needed to go from a room in section (x1,y1)(x1,y1) to a room in section (x2,y2)(x2,y2)?"
Input
The first line contains five integers n,m,cl,ce,vn,m,cl,ce,v (2≤n,m≤1082≤n,m≤108, 0≤cl,ce≤1050≤cl,ce≤105, 1≤cl+ce≤m−11≤cl+ce≤m−1, 1≤v≤n−11≤v≤n−1) — the number of floors and section on each floor, the number of stairs, the number of elevators and the maximum speed of an elevator, respectively.
The second line contains clcl integers l1,…,lcll1,…,lcl in increasing order (1≤li≤m1≤li≤m), denoting the positions of the stairs. If cl=0cl=0, the second line is empty.
The third line contains cece integers e1,…,ecee1,…,ece in increasing order, denoting the elevators positions in the same format. It is guaranteed that all integers lili and eiei are distinct.
The fourth line contains a single integer qq (1≤q≤1051≤q≤105) — the number of queries.
The next qq lines describe queries. Each of these lines contains four integers x1,y1,x2,y2x1,y1,x2,y2 (1≤x1,x2≤n1≤x1,x2≤n, 1≤y1,y2≤m1≤y1,y2≤m) — the coordinates of starting and finishing sections for the query. It is guaranteed that the starting and finishing sections are distinct. It is also guaranteed that these sections contain guest rooms, i. e. y1y1 and y2y2 are not among lili and eiei.
Output
Print qq integers, one per line — the answers for the queries.
Example
input
Copy
5 6 1 1 32531 1 5 61 3 5 43 3 5 3
output
Copy
754
Note
In the first query the optimal way is to go to the elevator in the 5-th section in four time units, use it to go to the fifth floor in two time units and go to the destination in one more time unit.
In the second query it is still optimal to use the elevator, but in the third query it is better to use the stairs in the section 2.
题意:
有n层楼，每层有m个单元，有的单元房有电梯，有的单元房有楼梯，楼梯移动一层需要1个时间单位
电梯一个时间单位移动v层，同一层相邻单元之间移动消耗1个时间单位
现在有q次查询
每次查询给出起点和终点的层数和单元位置
问从起点到终点最少需要几个时间单位
做法：
二分查找起点和终点之间的和偏左的横坐标的左、偏右的横坐标的右侧的电梯和楼梯的位置，取最优答案即可
叉点:
特判同一层，不需要找楼梯和电梯了，直接过去就好
代码：
1 #include<iostream>
2 using namespace std;
3 #include<cstdio>
4 #include<cstdlib>
5 #include<cstring>
6 #include<algorithm>
7 typedef long long ll;
8 typedef long long LL;
9 const ll maxn = 210000;
10 ll dt[maxn],lt[maxn];
11 ll n,m,cl,cd,v;
12 int binsearch_big(LL a[],int n,int s){
13 int mid;
14 LL l = 1,r = n;
15 while(l<=r){
16 mid = (l+r)/2;
17 if(a[mid]>=s)
18 r = mid - 1;
19 else
20 l = mid + 1;
21 }
22 if(l>=1&&l<=n)
23 return l;
24 else
25 return 0;
26 }
27 int binsearch_small(LL a[],int n,int d){
28 int mid;
29 LL l = 1,r = n;
30 while(l<=r){
31 mid = (l+r)/2;
32 if(a[mid]<=d)
33 l = mid + 1;
34 else
35 r = mid - 1;
36 }
37 if(r>=1&&r<=n)
38 return r;
39 else
40 return 0;
41 }
42 int main(){
43 scanf("%I64d%I64d%I64d%I64d%I64d",&n,&m,&cl,&cd,&v);
44 for(int i=1;i<=cl;i++){
45 scanf("%I64d",<[i]);
46 }
47 for(int i=1;i<=cd;i++){
48 scanf("%I64d",&dt[i]);
49 }
50 sort(lt+1,lt+cl+1);
51 sort(dt+1,dt+cd+1);
52 int sq;
53 scanf("%d",&sq);
54 for(int i=0;i<sq;i++){
55 LL x1,x2,y1,y2;
56 scanf("%I64d%I64d%I64d%I64d",&y1,&x1,&y2,&x2);
57 if(y1==y2){
58 LL ans = abs(x1-x2);
59 printf("%I64d\n",ans);
60 continue;
61 }
62 LL ansl = 0,ansd = 0;
63 ansl = abs(y2-y1);
64 ansd = abs(y2-y1)/v+(abs(y2-y1)%v!=0);
65 LL ll = x1,rr = x2;
66 if(ll>rr)
67 swap(ll,rr);
68 LL l = binsearch_big(lt,cl,ll);
69 LL r = binsearch_small(lt,cl,rr);
70 if((l>0&<[l]>=ll&<[l]<=rr)||(r>0&<[r]>=ll&<[r]<=rr)){
71 ansl+=abs(x2-x1);
72 }
73 else{
74 LL wr = binsearch_big(lt,cl,rr);
75 LL ans1 = 0x3f3f3f3f,ans2 = 0x3f3f3f3f;
76 if(wr>0)
77 ans2 = abs(ll-lt[wr])+abs(rr-lt[wr]);
78 LL wl = binsearch_small(lt,cl,ll);
79 if(wl>0)
80 ans1 = abs(ll-lt[wl])+abs(rr-lt[wl]);
81 ansl+=min(ans1,ans2);
82 }
83
84
85 l = binsearch_big(dt,cd,ll);
86 r = binsearch_small(dt,cd,rr);
87 if((l>0&&dt[l]<=rr&&dt[l]>=ll)||(r>0&&dt[r]>=ll&&dt[r]<=rr)){
88 ansd+=abs(x2-x1);
89 }
90 else{
91 LL wr = binsearch_big(dt,cd,rr);
92 LL ans1 = 0x3f3f3f3f,ans2 = 0x3f3f3f3f;
93 if(wr>0)
94 ans2 = abs(ll-dt[wr])+abs(rr-dt[wr]);
95 LL wl = binsearch_small(dt,cd,ll);
96 if(wl>0)
97 ans1 = abs(ll-dt[wl])+abs(rr-dt[wl]);
98 ansd+=min(ans1,ans2);
99 }
100 LL ans = min(ansd,ansl);
101 printf("%I64d\n",ans);
102 }
103 return 0;
104 }
原文地址：https://www.cnblogs.com/xfww/p/8973137.html