ACM: uva 1468 - Restaurant

Restaurant

Mr. Kim is planning to open a new restaurant. His city is laid
out as a grid with
size M x M.
Therefore, every road is horizontal or vertical and the horizontal
roads (resp., the vertical roads) are numbered from 0
to M - 1. For
profitability, all restaurants are located near road junctions. The
city has two big apartments which are located on the same
horizontal road. The figure below shows an example of a city map
with
size 11 x 11.
A circle represents an existing restaurant and a circle labeled
with `A' or `B' represents the location of an apartment. Notice
that a restaurant is already located at each apartment. Each road
junction is represented by the coordinate of the ordered pair of a
vertical road and a horizontal road. The distance between two
locations (x1, y1) and (x2, y2) is
computed
as | x1 - x2|
+
| y1 - y2|.
In the figure below, the coordinates of A and B
are (0, 5) and(10, 5),
respectively.





 
Mr. Kim knows that the residents of the two apartments
frequently have a meeting. So, he thinks that the best location of
a new restaurant is halfway between two apartments. Considering
lease expenses and existing restaurants, however, he can't select
the optimal location unconditionally. Hence he decides to regard a
location satisfying the following condition as
a good place.
Let dist(p, q) be
the distance
between p and q.

A location p is
a good place if for each existing
restaurant's
location q, dist(p, A)
<<I>dist(q, A) or dist(p, B)
< dist(q, B).
In other words, p is not a
good place if there exists an existing restaurant's
location q such
that dist(p, A)
dist(q, A) and dist(p, B)
dist(q, B).

In the above figure, the location (7,
4) is a good place. But the
location p = (4,
6) is not good because there is no apartment which
is closer to p than the
restaurant at q = (3, 5),
i.e., dist(p,A) = 5
dist(q, A) =
3 and dist(p, B)
= 7 dist(q, B) = 7. Also, the
location (0, 0) is not good due
to the restaurant at (0, 5). Notice that the
existing restaurants are positioned regardless of Mr. Kim's
condition.

Given n locations of
existing restaurants, write a program to compute the number of good
places for a new restaurant.

Input 

Your program is to read the input from standard input. The input
consists of T test cases.
The number of test
cases T is given in the
first line of the input. Each test case starts with a line
containing two
integers M and n ( 2
M 60, 000 and 2 n
50, 000), which represent the size of a city map and the number of
existing restaurants, respectively.
The (i + 1)-th line of a
test case contains two
integers xi and yi ( i =
1,
2,..., n and 0
xi, yi < M),
which represents the coordinate of the i-th
existing restaurant. Assume that all restaurants have distinct
coordinates and that the two apartments A and B are positioned at
the locations of 1-st restaurant and 2-nd restaurant. Notice that A
and B are placed on the same horizontal line.

Output 

Your program is to write to standard output. Print exactly one
line for each test case. Print the number of good places which can
be found in a given city map.

The following shows sample input and output for two test
cases.

Sample
Input 

2

6 3

1 3

4 3

0 2

11 11

0 5

10 5

4 9

2 8

7 8

5 6

3 5

5 3

3 2

7 2

9 1

Sample
Output 

2

16

题意:在一个M*M的矩形范围内, 有着A,B两间apartment, 有n-2间restaurant,
现在要你新建一件restaurant,

     要求你的restaurant要么比其它的restaurant靠近A,
要么比其它的restaurant靠近B. 距离是哈密顿距离.

    
要你计算出有多少个位置适合间这样的一间restaurant.

 

解题思路:

    
1. A,B在同一纵坐标, 横坐标的上下界是A,B的横坐标, 从左到右根据横坐标进行扫描, 找出每个横坐标

       
出现的大于A,B纵坐标的做小纵坐标和小于A,B纵坐标的最大纵坐标.(靠近AB直线).

    
2. 从两个方向一次扫描合法的区域, 统计合法位置数量即可.

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define min(a,b) (a)<(b)?(a):(b)

#define max(a,b) (a)>(b)?(a):(b)

#define MAX 60005

typedef long long ll;

const int INF = (1<<29);

int M, n;

int y[MAX], miny[MAX], maxy[MAX];

int xa, ya, xb, yb;

inline void swap(int &a, int
&b)

{

 int temp = a;

 a = b;

 b = temp;

}

int main()

{

// freopen("input.txt", "r", stdin);

 int caseNum, i;

 scanf("%d", &caseNum);

 while(caseNum--)

 {

  scanf("%d %d",
&M, &n);

  scanf("%d %d %d %d",
&xa, &ya, &xb,
&yb);

  if(xa > xb)
swap(xa, xb);

  for(i = xa+1; i
< xb; ++i)

  {

   miny[i] =
INF;

   maxy[i] =
-INF;

  }

  int tx, ty;

  for(i = 2; i <
n; ++i)

  {

   scanf("%d
%d", &tx, &ty);

   if(ty
>= ya) miny[tx] = min(ty, miny[tx]);

   if(ty
<= ya) maxy[tx] = max(ty, maxy[tx]);

  }

  y[xa] = 0;

  for(i = xa+1; i
< xb; ++i)

   y[i] =
min(miny[i]-ya, ya-maxy[i]);

  for(i = xa+1; i
< xb; ++i)

   y[i] =
min(y[i-1]+1, y[i]);

  y[xb] = 0;

  for(i = xb-1; i
> xa; --i)

   y[i] =
min(y[i+1]+1, y[i]);

  ll ans = 0;

  for(i = xa+1; i
< xb; ++i)

  {

   if(y[i])

   {

    ans++;

    ans
+= min(y[i]-1, M-1-ya);

    ans
+= min(y[i]-1, ya);

   }

  }

  printf("%lld\n", ans);

 }

 return 0;

}