2020牛客暑期多校训练营（第二场）B-Boundary

题目链接：
https://ac.nowcoder.com/acm/contest/5667/B
Given {n}n points in 2D plane. Considering all circles that the origin point {(0, 0)}(0,0) is on their boundries, find the one with the maximum given points on its boundry. Print the maximum number of points.

题意：给你若干个点，找一个过原点的圆使圆上经过的点最多。
思路：因为数据范围给的1e3，所以可以n^2地跑。由于三点确定一个圆，有一个点又是知道的(0,0)，那可以枚举其余每对点。找到他们分别与原点的中垂线的交点，continue掉中垂线平行的情况

比赛结束前6分钟才意识到map是log(n)的。。。此时已经WT了15发了。

正确处理：看了dl代码恍然大悟。可以用一个数组先把所有的点存下来，sort一下，连续出现一样的就是那个位置的统计次数。得到最大次数之后，由于是所有的点里面两两取的，出现了重复。枚举答案，计算谁的组合数是maxm，即为所求。

关于中垂线交点如何算比较方便，某考研选手建议解方程组用克拉默法则。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e6 + 100;
const int mod = 998244353;

vector<pair<int, int> > p;
vector<pair<double, double> > cur;

ll getD(ll a11, ll a12, ll a21, ll a22) {
    return a11 * a22 - a12 * a21;
}

int main() {
   //ios::sync_with_stdio(0);
    int n;
    p.resize(2005);
    scanf("%d", &n);
    int x, y;
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &x, &y);
        p[i] = make_pair(x, y);
    }

    for (int i = 1; i <= n - 1; i++) {
        for (int j = i + 1; j <= n; j++) {
            int x1 = p[i].first, y1 = p[i].second, x2 = p[j].first, y2 = p[j].second;
            ll a11 = 2 * x1, a12 = 2 * y1, a13 = x1 * x1 + y1 * y1;
            ll a21 = 2 * x2, a22 = 2 * y2, a23 = x2 * x2 + y2 * y2;
            ll D = getD(a11, a12, a21, a22);
            
            if (D == 0) {
                continue;
            }
            ll D1 = getD(a13, a12, a23, a22);
            ll D2 = getD(a11, a13, a21, a23);

            cur.push_back(make_pair((double)D1 / D, (double)D2 / D));
        }
    }

    sort(cur.begin(), cur.end());
    int mx = 0;
    for (int i = 0; i < cur.size(); ) {
        int j = i;
        while (j < cur.size() && cur[i] == cur[j]) j++;
        mx = max(mx, j - i);
        i = j;
    }
    int ans = 0;
    for (int u = 1; u < 5005; u++)
        if (u * (u - 1) / 2 <= mx)
            ans = max(ans, u);

    printf("%d\n", ans);

    return 0;
}