Power Transmission (Hard Edition)

大概:给n个点 每两点都有连线 求线段相交的次数

相交斜率一定不同 (斜率相同的条数)*(其他的) 就是于其他相交的次数 总数就/2

#include <bits/stdc++.h>
using namespace std;

int gcd(int a, int b) {
    return !b ? a : gcd(b, a % b);
}

int n, x[1005], y[1005];

struct line {
    int a, b, c;
    line(int x = 0, int y = 0, int z = 0) {
        int g = gcd(gcd(x, y), z);
        a = x / g, b = y / g, c = z / g;
    }
};
vector<line> vl;
bool operator<(const line &a, const line &b) {
    if (a.a != b.a)
        return a.a < b.a;
    if (a.b != b.b)
        return a.b < b.b;
    if (a.c != b.c)
        return a.c < b.c;
    return false;
}
bool operator==(const line &a, const line &b) {
    return a.a == b.a && a.b == b.b && a.c == b.c;
}
map<line, int> sz;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%d%d", x + i, y + i);
    for (int i = 1; i < n; ++i)
        for (int j = i + 1; j <= n; ++j)
            vl.emplace_back(y[i] - y[j], x[j] - x[i], x[i] * y[j] - x[j] * y[i]);
    sort(vl.begin(), vl.end());
    vl.erase(unique(vl.begin(), vl.end()), vl.end());
    int tot = vl.size();
    for (const line &p : vl)
        ++sz[line(p.a, p.b, 0)];
    long long ans = 0;
    for (const auto &p : sz)
        ans += 1LL * p.second * (tot - p.second);
    printf("%lld
", ans / 2);
    return 0;
}