题解

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct node{
	int x, y;
}a[1005];
bool cmp(node a, node b){
	if(a.x == b.x) return a.y < b.y;
	return a.x < b.x;
}bool check(node i, node j, node k){
	int ijx = i.x - j.x, ijy = i.y - j.y, jkx = j.x - k.x, jky = j.y - k.y;
	return ijx * jky == jkx * ijy;//转换距离比例成乘法
}
int main(){
	int n;
	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> a[i].x >> a[i].y;
	}
	int mx = 1;//如果只有一个点，不会循环，所以要把最开始初始化为1
	for(int i = 1; i <= n; i++){
		for(int j = i + 1; j <= n; j++){
			int ct = 0;
			for(int k = 1; k <= n; k++){
				if(check(a[i], a[j], a[k])) ct++;//如果三个点在一条直线上，计数
			}mx = max(mx, ct);
		}
	}
	cout << mx;
	
	return 0;
}

时间复杂度:O(n³)