POJ 3304 Segments(直线与线段相交)

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 1001;
const double eps = 1e-10;
int N;
struct Point {
    double x, y;
    Point operator - (const Point &rhs) const {
        return {x - rhs.x, y - rhs.y};
    }
    double operator ^ (const Point &rhs) const {
        return x * rhs.y - y * rhs.x;
    }
    bool operator == (const Point &rhs) const {
        return (x - rhs.x) < eps && (y -rhs.y) < eps;
    }
};
struct L {
    Point a, b;
};
Point line[MAXN][2];
int dcmp(double x) {
    if(fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}
bool judge(L l1, L l2) {
    Point tmp = l1.a - l1.b;
    bool flag = dcmp(dcmp(tmp ^ (l2.b - l1.b)) * dcmp(tmp ^ (l2.a - l1.b))) != 1;
    return flag;
}
bool check(Point a, Point b) {
    if(a == b) return 0;
    for(int i = 1; i <= N; i++)
        if(!judge({a, b}, {line[i][0], line[i][1]})) return 0;
    return 1;
}
void solve() {
    scanf("%d", &N);
    for(int i = 1; i <= N; i++) 
        scanf("%lf %lf %lf %lf", &line[i][0].x, &line[i][0].y, &line[i][1].x, &line[i][1].y);
    for(int i = 1; i <= N; i++)
        for(int j = 0; j < 2; j++)
            for(int k = 1; k <= N; k++)
                for(int l = 0; l < 2; l++)
                    if(check(line[i][j], line[k][l])) {puts("Yes!"); return ;}
    puts("No!");
}
int main() {
    int T; scanf("%d", &T);
    for(; T; T--, solve());
}