POJ 1269 Intersecting Lines(计算几何)

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给定两直线，判定相交，重合，或求出交点

验模板的题目

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

int t;

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) {
        this->x = x;
        this->y = y;
    }
    void read() {
        scanf("%lf%lf", &x, &y);
    }
};

typedef Point Vector;

Vector operator + (Vector A, Vector B) {
    return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Vector A, Vector B) {
    return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Vector A, double p) {
    return Vector(A.x * p, A.y * p);
}

Vector operator / (Vector A, double p) {
    return Vector(A.x / p, A.y / p);
}

const double eps = 1e-8;

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积

bool LineCoincide(Point p1, Point p2, Point p3) {
    return dcmp(Cross(p2 - p1, p3 - p1)) == 0;
}

bool LineParallel(Point P, Vector v, Point Q, Vector w) {
    return dcmp(v.x * w.y - v.y * w.x) == 0;
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
    Vector u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}

int main() {
    scanf("%d", &t);
    printf("INTERSECTING LINES OUTPUT\n");
    while (t--) {
        Point p1, p2, p3, p4;
        p1.read(); p2.read();
        p3.read(); p4.read();
        if (LineCoincide(p1, p2, p3) && LineCoincide(p1, p2, p4))
            printf("LINE\n");
        else if (LineParallel(p1, p2 - p1, p3, p4 - p3))
            printf("NONE\n");
        else {
            Point ans = GetLineIntersection(p1, p2 - p1, p3, p4 - p3);
            printf("POINT %.2f %.2f\n", ans.x, ans.y);
        }
    }
    printf("END OF OUTPUT\n");
    return 0;
}

POJ 1269 Intersecting Lines(计算几何)

标签：

原文地址：http://blog.csdn.net/accelerator_/article/details/44080923