给定一个平面内的点的集合，求共线最多点的个数。要求减少重复运算，并给出测试案例。

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思路：对于集合中的任意两点，循环遍历剩下的点，判断是否与其共线，记录下共线点的个数，找出其中一个最大的共线点的个数

#define _CRT_SECURE_NO_WARNINGS#include <iostream>  #include <string>  #include <algorithm>   #include <vector>  #include <ctime>  using namespace std;struct Point {int x;int y;Point() : x(0), y(0) {}Point(int a, int b) : x(a), y(b) {}};class pointSolution {public:bool isCollineation(Point& p1, Point& p2, Point& p3)//判断三点是否共线{int x1 = p2.x - p1.x;int y1 = p2.y - p1.y;int x2 = p3.x - p1.x;int y2 = p3.y - p1.y;if (x1 * y2 == x2 * y1)  //共线的定义return true;return false;}int maxPoints(vector<Point> &points){if (points.size() <= 2){return points.size();}int maxLine = 2;bool flag = false;//检测是否所有点都重合  for (size_t i = 0; i < points.size(); ++i){int repNum = 0;//记录重复点的个数  for (size_t j = i + 1; j < points.size(); ++j){if (points[i].x == points[j].x && points[i].y == points[j].y){repNum++;continue;}flag = true;int count = 2 + repNum;for (size_t k = j + 1; k < points.size(); ++k)if (isCollineation(points[i], points[j], points[k]))count++;if (count > maxLine)maxLine = count;}}if (!flag){maxLine = points.size();}return maxLine;}};int main(){srand((unsigned int) time(NULL));int x = 0, y = 0;vector<Point> points;for(int i = 0; i < 50; i++){points.push_back(Point(rand() % 100, rand() % 100));}pointSolution ps;cout <<"共线最多点的个数是："<< ps.maxPoints(points) << endl;return 0;}

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