牛客多校第二场 B Boundary

牛客多校第二场 B

链接：https://ac.nowcoder.com/acm/contest/5667/B

题目描述

Given (n) points in 2D plane. Considering all circles that the origin point ((0, 0)) is on their boundries, find the one with the maximum given points on its boundry. Print the maximum number of points.

输入描述:

The first line contains one integer (n~(1 leq n leq 2000)) , denoting the number of given points.
Following (n) lines each contains two integers (x, y~(|x|,|y| leq 10000)), denoting a given point ((x, y)).
It's guaranteed that the points are pairwise different and no given point is the origin point.

输出描述:

Only one line containing one integer, denoting the answer.

输入

4
1 1
0 2
2 0
2 2

输出

3

说明

Considering circle ((x-1)^2+(y-1)^2=2), we can see that the origin point is on its boundry and that there are (3) given points ({(0,2),(2,0),(2,2)}) on its boundry.

题解

题意

在一个二维平面中，给 (n) 个点，以 ((x,y)) 为圆心的圆经过 ((0,0)) 点，问，这个圆在这 (n) 个点中最多能经过多少个点。

思路

对于两个不同的点 ((a_i,b_i))，((a_j,b_j)) ，还有额外加的一个 ((0,0)) 点在三点不共线的条件下，一定可以确定一个圆，并且可以求出圆心，

[(x-a_i)^2+(y-b_i)^2=x^2+y^2 \ (x-a_j)^2+(y-b_j)^2=x^2+y^2 \ (x-a_i)^2+(y-b_i)^2=(x-a_j)^2+(y-b_j)^2 ]

化简式子，可以得到一个关于 $x,y $ 的方程组

[y=a_j(a_i^2+b_i^2)-a_i(a_j^2+b_j^2)/(2(b_ia_j-a_ib_j)) \ y=b_j(a_i^2+b_i^2)-b_i(a_j^2+b_j^2)/(2(b_ia_j-a_ib_j)) ]

之后进行求解储存数据即可。

注意

• (map) 会卡 (tle)
• 卡精度，能不使用 (double) 尽量不要使用 (double)
• 当三点共线时，不进行考虑

代码

#pragma GCC optimize(3)
#include <cmath>
#include <vector>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 2e3+55;
const double eps = 1e-7;
struct Point{double x,y;};
bool cmp(Point x,Point y){if(eps>fabs(x.x-y.x))return x.y<y.y;return x.x<y.x;}
struct node{ll a,b;ll aa,bb;}nn[maxn];
vector<Point>v;
int main(){
    int n;
    int maxxx=0;
    scanf("%d",&n);
    if(n==1){printf("1
");return 0;}
    for(int i=0;i<n;++i){
        v.clear();
        scanf("%lld%lld",&nn[i].a,&nn[i].b);
        nn[i].aa=nn[i].a*nn[i].a;
        nn[i].bb=nn[i].b*nn[i].b;
        for(int j=0;j<i;++j){
            if(nn[i].b*nn[j].a==nn[j].b*nn[i].a)continue;
            ll yx=nn[j].a*(nn[i].aa+nn[i].bb)-nn[i].a*(nn[j].aa+nn[j].bb);
            ll yy=(ll)2*(nn[i].b*nn[j].a-nn[j].b*nn[i].a);
            ll xx=nn[j].b*(nn[i].aa+nn[i].bb)-nn[i].b*(nn[j].aa+nn[j].bb);
            ll xy=-yy;
            v.push_back({(double)xx/xy,(double)yx/yy});
        }
        int maxx=0,ans=1;
        sort(v.begin(),v.end(),cmp);
        for(int j=1;j<(int)v.size();++j){
            if(fabs(v[j].y-v[j-1].y)<eps&&fabs(v[j].x-v[j-1].x)<eps)ans++;
            else ans=1;
            maxx=max(maxx,ans);
        }
        maxxx=max(maxxx,maxx+1);
    }
    printf("%d
",maxxx);
    return 0;
}