题目链接:https://abc091.contest.atcoder.jp/tasks/arc092_a 题意 On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (ai,bi), and the coordinates of the i-th blue point are (ci,di). A red point and…
Problem Statement On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (ai,bi), and the coordinates of the i-th blue point are (ci,di). A red point and a blue point can form a friendly pair w…
Problem Statement On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (ai,bi), and the coordinates of the i-th blue point are (ci,di). A red point and a blue point can form a friendly pair w…
Find the K closest points to the origin in a 2D plane, given an array containing N points. 用 max heap 做 /* public class Point { public int x; public int y; public Point(int x, int y) { this.x = x; this.y = y; } } */ public List<Point> findKClosest(P…
C - 2D Plane 2N Points 把能连边的点找到然后跑二分图匹配即可 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define space putchar(' ') #define enter putchar('\n') #define mp make_pair #define MAXN 100005 //#define ivorysi u…
After the war, the supersonic rocket became the most common public transportation. Each supersonic rocket consists of two "engines". Each engine is a set of "power sources". The first engine has nn power sources, and the second one has…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 这道题给了我们一堆二维点,然后让我们求最大的共线点的个数,根据初中数学我们知道,两点确定一条直线,而且可以写成y = ax + b的形式,所有共线的点都满足这个公式.所以这些给定点两两之间都可以算一个斜率,每个斜率代表一条直线,对每一条直线,带入所有的点看是否共线并计算个数,这是整体的思路.但是还有…
题目链接 Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 分析:首先要注意的是,输入数组中可能有重复的点.由于两点确定一条直线,一个很直观的解法是计算每两个点形成的直线,然后把相同的直线合并,最后包含点最多的直线上点的个数就是本题的解.我们知道表示一条直线可以用斜率和y截距两个浮点数(垂直于x轴的直线斜率为无穷大,截距用x截距),同时还需要保存每…
Find the K closest points to a target point in a 2D plane. class Point { public int x; public int y; public Point(int x, int y) { this.x = x; this.y = y; } } class Solution { public List<Point> findKClosest(Point[] points, int k, Point p) { // max h…
Max Points on a Line 题目描述: Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 解题思路: 1.首先由这么一个O(n^3)的方法,也就是算出每条线的方程(n^2),然后判断有多少点在每条线上(N).这个方法肯定是可行的,只是复杂度太高2.然后想到一个O(N)的,对每一个点,分别计算这个点和其他所有点构成的斜率,具有相同斜率最…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 思路 关键是浮点数做key不靠谱,struct hash以及 int calcGCD(int a, int b)的写法 代码 /** * Definition for a point. * struct Point { * int x; * int y; * Point() : x(0), y(0)…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 思路: 自己脑子当机了,总是想着斜率和截距都要相同.但实际上三个点是一条直线的话只要它们的斜率相同就可以了,因为用了相同的参照点,截距一定是相同的. 大神的做法: 对每一个点a, 找出所有其他点跟a的连线斜率,相同为同一条线,记录下通过a的点的线上最大的点数. 找出每一个点的最大连线通过的点数. 其…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 主要思想:O(n2),固定一个点,遍历其余 n 个点, 计算与该点相同的点的个数,和其余所有点的斜率,相同斜率的点视为同一直线. 初步 AC 代码: /** * Definition for a point. * struct Point { * int x; * int y; * Point()…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 解题思路: 本题主要需要考虑到斜线的情况,可以分别计算出过points[i]直线最多含几个点,然后算出最大即可,由于计算points[i]的时候,前面的点都计算过了,所以不需要把前面的点考虑进去,所以问题可以转化为过points[i]的直线最大点的个数,解题思路是用一个HashMap储存斜率,遍历p…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. /** * Definition for a point. * struct Point { * int x; * int y; * Point() : x(0), y(0) {} * Point(int a, int b) : x(a), y(b) {} * }; */ class Solutio…
Problem: Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Suppose that the structure Point is already defined in as following: /** * Definition for a point. * struct Point { * int x; * int y; * Point…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 求二维平面上n个点中,最多共线的点数. 1.比较直观的方法是,三层循环,以任意两点划线,判断第三个点是否在这条直线上. 比较暴力 2.使用map来记录每个点的最大数目. /** * Definition for a point. * class Point { * int x;…
题目: Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 链接: http://leetcode.com/problems/max-points-on-a-line/ 题解: 这道题是旧时代的残党,LeetCode大规模加新题之前的最后一题,新时代没有可以载你的船.要速度解决此题,之后继续刷新题. 主要思路是,对每一个点求其于其他点的斜率,放在一个…
1.问题描述 Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. 2.翻译 对于在一个平面上的N个点,找出在同一条直线的最多的点的数目 3.思路分析 我们知道任意的两个点可以构成一条直线,对于一条在直线的上的点,他们必然具有相同的斜率,这个我初中都知道了.因为找出最多的点的方法也就比较简单了,我们只要依次遍历这些点,并且记录相同的斜率的点的数目,这样…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. /** * Definition for a point. * struct Point { * int x; * int y; * Point() : x(0), y(0) {} * Point(int a, int b) : x(a), y(b) {} * }; */ class Solutio…
OpenCASCADE BRepMesh - 2D Delaunay Triangulation eryar@163.com Abstract. OpenCASCADE package BRepMesh can compute the Delaunay’s triangulation with the algorithm of Watson. It can be used for 2d plane or on surface by meshing in UV parametric space.…
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Example 1: Input: [[1,1],[2,2],[3,3]] Output: 3 Explanation: ^ | | o | o | o +-------------> 0 1 2 3 4 Example 2: Input: [[1,1],[3,2]…
