codeforces A. Supercentral Point 题解-优快云博客

博客围绕在笛卡尔坐标系中计算超中心点数量展开。先定义了点的邻居及超中心点概念，给出输入输出要求。提到暴力法可解决问题，效率为O(n^2)，还指出使用hash表能将效率近乎降到O(n)，需巧妙结合map和set容器。

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
8
1 1
4 2
3 1
1 2
0 2
0 1
1 0
1 3

output
2

暴力法可过，效率O(n^2)

可是使用hash表能够把效率降到近乎O(n)

要巧妙使用两个map容器。

要对map和set容器非常熟悉了。合起来一起使用。

#include <unordered_map>
#include <set>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
using namespace std;

void SupercentralPoint()
{
	int n = 0, x, y;
	cin>>n;
	unordered_map<int, set<int> > xumIS;
	unordered_map<int, set<int> > yumIS;
	pair<int, int> *pii = new pair<int, int>[n];
	for (int i = 0; i < n; i++)
	{
		cin>>x>>y;
		pii[i].first = x;
		pii[i].second = y;
		xumIS[x].insert(y);
		yumIS[y].insert(x);
	}
	int supers = 0;
	for (int i = 0; i < n; i++)
	{
		if ( pii[i].second != *(xumIS[pii[i].first].begin()) &&
			pii[i].second != *(xumIS[pii[i].first].rbegin()) &&
			pii[i].first != *(yumIS[pii[i].second].begin()) &&
			pii[i].first != *(yumIS[pii[i].second].rbegin()) )
			supers++;
	}
	cout<<supers;
	delete [] pii;
}

转载于:https://www.cnblogs.com/mqxnongmin/p/10558491.html