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

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

The guests can walk along the corridor on each floor, use stairs and elevators. Each stairs or elevator occupies all sections $(1,x)(1,x), (2,x)(2,x), \dots \dots , (n,x)(n,x) for\; some xx between 11 and mm.\; All\; sections\; not\; occupied\; with\; stairs\; or\; elevators\; contain\; guest\; rooms.\; It\; takes\; one\; time\; unit\; to\; move\; between\; neighboring\; sections\; on\; the\; same\; floor\; or\; to\; move\; one\; floor\; up\; or\; down\; using\; stairs.\; It\; takes\; one\; time\; unit\; to\; move\; up\; to vvfloors\; in\; any\; direction\; using\; an\; elevator.\; You\; can\; assume\; you\; don\text{'}t\; have\; to\; wait\; for\; an\; elevator,\; and\; the\; time\; needed\; to\; enter\; or\; exit\; an\; elevator\; is\; negligible.$

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

Input

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

Output

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

Example

input

Copy

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

output

Copy

7
5
4

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.

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 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",&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 }