Codeforces Round #112 (Div. 2)---A. Supercentral Point

Supercentral Point

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):

• point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
• point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
• point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
• point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y

We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.

Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set.

Input

The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different.

Output

Print the only number — the number of supercentral points of the given set.

Sample test(s)

input

81 14 23 11 20 20 11 01 3

output

2

input

50 00 11 00 -1-1 0

output

1

Note

In the first sample the supercentral points are only points (1, 1) and (1, 2).

In the second sample there is one supercental point — point (0, 0).

解题思路：没什么说的，直接暴力搞了。遍历每个点，看是否符合要求。为了省时间，我们可以在输入的时候把x的上限,下限，和y的上限和下限先记录一下，在判断每个点的时候会用到。

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint x[205], y[205], a[2005][2005];int main(){    #ifdef sxk        freopen("in.txt","r",stdin);    #endif    int n, xx, yy, xxx, yyy, flag0, flag1, flag2, flag3;    while(scanf("%d",&n)!=EOF)    {        memset(a, 0, sizeof(a));        xx = yy = -12345;        xxx= yyy = 12345;        for(int i=0; i<n; i++){            scanf("%d%d", &x[i], &y[i]);            x[i] += 1000;  y[i] += 1000;            a[x[i]][y[i]] = 1;            if(xx < x[i]) xx = x[i];       //纪录x，y范围            if(xxx > x[i]) xxx = x[i];            if(yy < y[i])  yy = y[i];            if(yyy > y[i]) yyy = y[i];        }        int ans = 0;        for(int i=0; i<n; i++){            flag0 = flag1 = flag2 = flag3 = 0;            for(int j=x[i]+1; j<=xx; j++){        //判断                if( a[j][ y[i] ] ){                    flag0 = 1;                    break;                }            }              if(flag0){                                        for(int j=xxx; j<x[i]; j++){                    if( a[j][ y[i] ] ){                        flag1 = 1;                        break;                    }                }                if(flag1){                    for(int j=y[i]+1; j<=yy; j++){                        if( a[x[i]][j] ){                            flag2 = 1;                            break;                        }                    }                    if(flag2){                        for(int j=yyy; j<y[i]; j++){                            if( a[x[i]][j] ){                                flag3 = 1;                                break;                            }                        }                    }                }            }            if(flag3) ans ++;        }        printf("%d\n", ans);    }    return 0;}