200652: [AtCoder]ARC065 E - Manhattan Compass

Memory Limit:256 MB Time Limit:3 S
Judge Style:Text Compare Creator:
Submit:0 Solved:0

Description

Score : $900$ points

Problem Statement

There are $N$ pinholes on the $xy$-plane. The $i$-th pinhole is located at $(x_i,y_i)$.

We will denote the Manhattan distance between the $i$-th and $j$-th pinholes as $d(i,j)(=|x_i-x_j|+|y_i-y_j|)$.

You have a peculiar pair of compasses, called Manhattan Compass. This instrument always points at two of the pinholes. The two legs of the compass are indistinguishable, thus we do not distinguish the following two states: the state where the compass points at the $p$-th and $q$-th pinholes, and the state where it points at the $q$-th and $p$-th pinholes.

When the compass points at the $p$-th and $q$-th pinholes and $d(p,q)=d(p,r)$, one of the legs can be moved so that the compass will point at the $p$-th and $r$-th pinholes.

Initially, the compass points at the $a$-th and $b$-th pinholes. Find the number of the pairs of pinholes that can be pointed by the compass.

Constraints

  • $2≦N≦10^5$
  • $1≦x_i, y_i≦10^9$
  • $1≦a < b≦N$
  • When $i ≠ j$, $(x_i, y_i) ≠ (x_j, y_j)$
  • $x_i$ and $y_i$ are integers.

Input

The input is given from Standard Input in the following format:

$N$ $a$ $b$
$x_1$ $y_1$
:
$x_N$ $y_N$

Output

Print the number of the pairs of pinholes that can be pointed by the compass.


Sample Input 1

5 1 2
1 1
4 3
6 1
5 5
4 8

Sample Output 1

4

Initially, the compass points at the first and second pinholes.

Since $d(1,2) = d(1,3)$, the compass can be moved so that it will point at the first and third pinholes.

Since $d(1,3) = d(3,4)$, the compass can also point at the third and fourth pinholes.

Since $d(1,2) = d(2,5)$, the compass can also point at the second and fifth pinholes.

No other pairs of pinholes can be pointed by the compass, thus the answer is $4$.


Sample Input 2

6 2 3
1 3
5 3
3 5
8 4
4 7
2 5

Sample Output 2

4

Sample Input 3

8 1 2
1 5
4 3
8 2
4 7
8 8
3 3
6 6
4 8

Sample Output 3

7

Input

题意翻译

在一个平面直角坐标系上有 $n$ 个点,第 $i$ 个点的坐标是 $(x_i,y_i)$,两个点 $i,j$ 之间的曼哈顿距离定义为 $dis(i,j)=|x_i-x_j|+|y_i-y_j|$。 现在有一个指南针指向了这 $n$ 个点中的两个点 $P(a,b)$,如果有一个点 $x$ 满足 $dis(a,x)=dis(a,b)$ 那么指南针的状态可以变更为 $P(a,x)$,否则如果点 $x$ 满足 $dis(b,x)=dis(a,b)$,那么指南针的状态可以变更为 $P(b,x)$。(注意 $P(i,j)$ 和 $P(j,i)$ 是同一种状态)。 现在给定 $n$,$a$,$b$,问你指南针总共有多少种不同的状态,即指南针能指向多少对点对。

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