In the year of $\mathrm{30XX30XX\; participants\; of\; some\; world\; programming\; championship\; live\; in\; a\; single\; large\; hotel.\; The\; hotel\; has\; nn\; floors.\; Each\; floor\; has\; mm\; sections\; with\; a\; single\; corridor\; connecting\; all\; of\; them.\; The\; sections\; are\; enumerated\; from\; 11\; to\; mm\; along\; the\; corridor,\; and\; all\; sections\; with\; equal\; numbers\; on\; different\; floors\; are\; located\; exactly\; one\; above\; the\; other.\; Thus,\; the\; hotel\; can\; be\; represented\; as\; a\; rectangle\; of\; height}$

The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections $(1,x)(1,x),\; (2,x)(2,x),\; \dots \dots ,$

You are to process $\mathrm{qq\; queries.\; Each\; query\; is\; a\; question\; "what\; is\; the\; minimum\; time\; needed\; to\; go\; from\; a\; room\; in\; section\; (x1,y1)(x1,y1)\; to\; a\; room\; in\; section\; (x2,y2)(x2,y2)?"}$

The first line contains five integers $\mathrm{n,m,cl,ce,vn,m,cl,ce,v\; (2\le n,m\le 1082\le n,m\le 108,\; 0\le cl,ce\le 1050\le cl,ce\le 105,\; 1\le cl+ce\le m-11\le cl+ce\le m-1,\; 1\le v\le n-11\le v\le n-1)\; —\; the\; number\; of\; floors\; and\; section\; on\; each\; floor,\; the\; number\; of\; stairs,\; the\; number\; of\; elevators\; and\; the\; maximum\; speed\; of\; an\; elevator,\; respectively.}$

The second line contains ${\mathrm{clcl\; integers\; l1,\dots ,lcll1,\dots ,lcl\; in\; increasing\; order\; (1\le li\le m1\le li\le m),\; denoting\; the\; positions\; of\; the\; stairs.\; If\; cl=0cl=0,\; the\; second\; line\; is\; empty.}}^{}$

The third line contains ${\mathrm{cece\; integers\; e1,\dots ,ecee1,\dots ,ece\; in\; increasing\; order,\; denoting\; the\; elevators\; positions\; in\; the\; same\; format.\; It\; is\; guaranteed\; that\; all\; integers\; lili\; and\; eiei\; are\; distinct.}}^{}$

The fourth line contains a single integer $\mathrm{qq\; (1\le q\le 1051\le q\le 105)\; —\; the\; number\; of\; queries.}$

The next $\mathrm{qq\; lines\; describe\; queries.\; Each\; of\; these\; lines\; contains\; four\; integers\; x1,y1,x2,y2x1,y1,x2,y2\; (1\le x1,x2\le n1\le x1,x2\le n,\; 1\le y1,y2\le m1\le y1,y2\le m)\; —\; the\; coordinates\; of\; starting\; and\; finishing\; sections\; for\; the\; query.\; It\; is\; guaranteed\; that\; the\; starting\; and\; finishing\; sections\; are\; distinct.\; It\; is\; also\; guaranteed\; that\; these\; sections\; contain\; guest\; rooms,\; i.\; e.\; y1y1\; and\; y2y2\; are\; not\; among\; lili\; and\; eiei.}$

Print $\mathrm{qq\; integers,\; one\; per\; line\; —\; the\; answers\; for\; the\; queries.}$

5 6 1 1 3
2
5
3
1 1 5 6
1 3 5 4
3 3 5 3

7
5
4

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.

題意:

有n層樓，每層有m個單元，有的單元房有電梯，有的單元房有樓梯，樓梯移動一層需要1個時間單位

電梯一個時間單位移動v層，同一層相鄰單元之間移動消耗1個時間單位

現在有q次查詢

每次查詢給出起點和終點的層數和單元位置

問從起點到終點最少需要幾個時間單位

做法：

二分查找起點和終點之間的和偏左的橫坐標的左、偏右的橫坐標的右側的電梯和樓梯的位置，取最優答案即可

叉點:

特判同一層，不需要找樓梯和電梯了，直接過去就好

代碼：

  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",&lt[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&&lt[l]>=ll&&lt[l]<=rr)||(r>0&&lt[r]>=ll&&lt[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 }

Codeforces Round #477 (rated, Div. 2, based on VK Cup 2018 Round 3) C. Stairs and Elevators【二分查找】