Smallest Rectangle

题目1 : Smallest Rectangle

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

You are given N 2D points P1, P2, ... PN. If there is a rectangle satisfying that its 4 vertices are in the given point set and its 4 sides are parallel to the axis we say the rectange exists.

Find the smallest exsisting rectange and output its area.

输入

The first line contains an integer N. (1 <= N <= 1000)

The following N lines each contain 2 integers Xi and Yi denoting a point (Xi, Yi). (0 <= Xi, Yi <= 1000000)

输出

Output the smallest area. If no rectangle exsists output -1.

样例输入

9  
0 0  
0 1   
0 4  
1 0  
1 1  
1 4  
4 0  
4 1  
4 4

样例输出

1

把所有的点都放在哈希表里，这样可以O(1)判断一个坐标(x, y)上是不是有点。

然后枚举其中两个点，不妨设是(Xi, Yi)和(Xj, Yj)，当作矩形的左上角顶点和右下角顶点。

这样我们只需判断(Xi, Yj)和(Xj, Yi)上有没有点即可判断是否构成一个矩形。如果构成矩形的化，更新当前最小面积。

总的时间复杂度是O(N^2)。

import java.util.*;
public class Main {
	static class Point implements Comparable<Point> {
		long x;
		long y;
		Point(long x, long y) {
			this.x = x;
			this.y = y;
		}

		public int hashCode()
		{
			return (int)((x << 20) + y); 
		}
	 	
	 	public boolean equals(Object obj)  
   	 	{
   	 		Point p = (Point)obj;  
        	return x == p.x && y == p.y;
    	} 

		public int compareTo(Point o) {
			if (x == o.x) {
				if (y == o.y) {
					return 0;
				} 
				return y < o.y ? -1 : 1;
			}
			return x < o.x ? -1 : 1;
		}
	}

	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		HashSet<Point> set = new HashSet<>();
		ArrayList<Point> arr = new ArrayList<>();
		for(int i = 0; i < n; i++) {
			long x = sc.nextLong();
			long y = sc.nextLong();
			arr.add(new Point(x, y));
			set.add(new Point(x, y));
		}

		Long ans = -1L;
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				long x1 = arr.get(i).x;
				long y1 = arr.get(i).y;
				long x2 = arr.get(j).x;
				long y2 = arr.get(j).y;
				if (x1 == x2 || y1 == y2) {
					continue;
				}
				if(set.contains(new Point(x1, y2)) && set.contains(new Point(x2, y1))) {
					long area = Math.abs((x1 - x2) * (y1 - y2));
					if (ans == -1L || area < ans) {
						ans = area;
					}
				}
			}
		}
		System.out.println(ans);
	}
}