cf1163C2

题意简述：给出N个点，任意两点之间有一条直线，问共有多少对直线相交？（N<=2000)

官方代码如下

#include <cstdio>
#include <map>
#include <set>
#include <utility>
const int N = 1001;
int x[N], y[N];
std::map<std::pair<int,int>,std::set<long long>> slope_map;

int gcd(int a, int b) 
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

int main()
{
    int n; scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%d%d", &x[i], &y[i]);
    long long total = 0, res = 0;
    for (int i = 1; i <= n - 1; ++i)
        for (int j = i + 1; j <= n; ++j)
        {
            int x1 = x[i], y1 = y[i], x2 = x[j], y2 = y[j];
            // construct a line passing through (x1, y1) and (x2, y2)
            // line is expressed as equation ax - by = c with constant a, b, c 
            int a = y1 - y2, b = x1 - x2;
            // simplify equation
            int d = gcd(a, b); a /= d, b /= d;
            if (a < 0 || (a == 0 && b < 0))
            {
                a = -a;
                b = -b;
            }
            // lines with the same slope (same a, b) are stored in a map
            std::pair<int,int> slope(a, b);
            long long c = 1LL * a * x1 - 1LL * b * y1;
            if (!slope_map[slope].count(c))
            {
                ++total;
                slope_map[slope].insert(c);
                // if this line is new, it intersects every line with different slope
                res += total - slope_map[slope].size();
            }
        }
    printf("%lld
", res);
}