Aizu - CGL_2_C Cross Point(计算几何，线段交点)

Description

For given two segments s1 and s2, print the coordinate of the cross point of them.

s1 is formed by end points p0 and p1, and s2 is formed by end points p2 and p3.

Input

The entire input looks like:

q (the number of queries)
1st query
2nd query
...
qth query

Each query consists of integer coordinates of end points of s1 and s2 in the following format:

xp0 yp0 xp1 yp1 xp2 yp2 xp3 yp3

Output

For each query, print the coordinate of the cross point. The output values should be in a decimal fraction with an error less than 0.00000001.

Constrains

• 1 ≤ q ≤ 1000
• -10000 ≤ xpi, ypi ≤ 10000
• p0 ≠ p1 and p2 ≠ p3.
• The given segments have a cross point and are not in parallel.

Sample Input

3
0 0 2 0 1 1 1 -1
0 0 1 1 0 1 1 0
0 0 1 1 1 0 0 1

Sample Output

1.0000000000 0.0000000000
0.5000000000 0.5000000000
0.5000000000 0.5000000000

---

#include <bits/stdc++.h>

using namespace std;

const double EPS = 1e-7;
inline bool equals(double a, double b) { return fabs(a - b) < EPS; }

class Point
{
public:
    double x, y;
    Point(double x = 0, double y = 0): x(x), y(y) {}

    Point operator+(Point p) { return Point(x + p.x, y + p.y); }
    Point operator-(Point p) { return Point(x - p.x, y - p.y); }
    Point operator*(double k) { return Point(x * k, y * k); }
    Point operator/(double k) { return Point(x / k, y / k); }

    double abs() { return sqrt(norm()); }
    double norm() { return x * x + y * y; }

    bool operator<(const Point & p) const { return x != p.x ? x < p.x : y < p.y; }

    bool operator==(const Point & p) const 
    { return fabs(x - p.x) < EPS && fabs(y - p.y) < EPS; }
};

typedef Point Vector;

struct Segment {
    Point p1, p2;  
};

double cross(const Vector & a, const Vector & b)
{
    return a.x * b.y - a.y * b.x;
}

double dot(const Vector & a, const Vector & b)
{
    return a.x * b.x + a.y * b.y;
}

int ccw(Point & p0, Point & p1, Point & p2)
{
    Vector a = p1 - p0;
    Vector b = p2 - p0;
    if (cross(a, b) > EPS) return -1;
    else if (cross(a, b) < -EPS) return 1;
    else {
        if (dot(a, b) < -EPS) return 2;
        else if (a.norm() < b.norm()) return -2;
        else return 0;
    }
}

bool intersection(Point & p1, Point & p2, Point & p3, Point & p4)
{
    return ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 
        && ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0;
}

Point getCrossPoint(Segment s1, Segment s2)
{
    Vector base = s2.p2 - s2.p1;
    double d1 = abs(cross(base, s1.p1 - s2.p1));
    double d2 = abs(cross(base, s1.p2 - s2.p1));
    double t = d1 / (d1 + d2);
    // Point tmp = s1.p2 - s1.p1;
    // tmp = tmp * t;
    // return s1.p1 + tmp;
    return s1.p1 + (s1.p2 - s1.p1) * t;
}

int main()
{   
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int caseNum; cin >> caseNum;
    while (caseNum--) {
        double x0, y0, x1, y1, x2, y2, x3, y3;
        cin >> x0 >> y0 >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
        Segment s1, s2;
        s1.p1 = Point(x0, y0); s1.p2 = Point(x1, y1);
        s2.p1 = Point(x2, y2); s2.p2 = Point(x3, y3);
        Point res = getCrossPoint(s1, s2);
        cout << fixed << setprecision(8) << res.x << ' ' << fixed << setprecision(8) << res.y << endl;
    }   

    return 0;
}

可以和LeetCode面试题16.03. 交点结合起来看，注意线段部分重合的情况。