Codeforces 1142C（转化、凸包）

可以变换坐标：x' = x, y' = y - x ^ 2，如此之后可得线性函数x' * b + c = y'，可以发现两点连边为抛物线，而其他点都在这条线下方才满足题意，故而求一个上凸壳即可。

#include <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

typedef long long ll;
const int maxn = 1e5 + 5;
int n;
struct Point {
    ll x, y;

    Point(){}
    Point(ll x, ll y):x(x), y(y) {}

    friend ll operator * (const Point &a, const Point &b) {
        return a.y * b.x - b.y * a.x;
    }

    friend Point operator - (const Point &a, const Point &b) {
        return Point(a.x - b.x, a.y - b.y);
    }

    bool operator < (const Point &rhs) const {
        if (x != rhs.x)    return x < rhs.x;
        return y < rhs.y;
    }
};

vector<Point> v, st;

void read() {
    cin >> n;

    for (int i = 0; i < n; i++) {
        ll x, y;
        cin >> x >> y;
        y = y - x * x;
        v.push_back({x, y});
    }
}

void convex() {
    sort(v.begin(), v.end());//按坐标排序

    for (int i = n - 1; ~i; --i)
        v[i] = v[i] - v[0];

    for (int i = 0; i < n; i++) {
        int tot = st.size();
        while (tot > 1 && (v[i] - st[tot - 2]) * (st[tot - 1] - st[tot - 2]) >= 0)    tot--, st.pop_back();
        st.push_back(v[i]);
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    read();
    convex();

    int ans = 0;
    for (int i = 1; i < st.size(); i++)
        if (st[i].x != st[i - 1].x)//x相同无法构成二次函数
            ans++;
    cout << ans << endl;

    return 0;
}