UVa 11722 (概率 数形结合) Joining with Friend

高中也做个这种类似的题目，概率空间是[t1, t2] × [s1, s2]的矩形，设x、y分别代表两辆列车到达的时间，则两人相遇的条件就是|x - y| <= w

从图形上看就是矩形夹在两条平行线之间的部分。

因为情况众多，一个一个分类很麻烦，而且可能有漏掉情况，所以就用计算几何的办法求了个凸多边形，多边形 与 矩形面积之比就是概率。

代码有点挫，将就看，=_=||

#include <cstdio>
#include <vector>
#include <cmath>
using namespace std;

struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x), y(y) {}
};
typedef Point Vector;
typedef vector<Point> Polygon;

Point operator + (const Point& A, const Point& B)
{ return  Point(A.x+B.x, A.y+B.y); }

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

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

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

double PolygonArea(const Polygon& p)
{//求多边形面积
     int n = p.size();
     double area = 0;
     for(int i = 1; i < n-1; i++)
        area += Cross(p[i]-p[0], p[i+1]-p[0]);
    return area/2;
}

Point Intersection(Point P, Vector v, Point Q, Vector w)
{//求两直线交点
    Vector u = P-Q;
    double t = Cross(w, u) / Cross(v, w);
    return P+v*t;
}

const double sqrt2 = sqrt(2.0);
const Vector v1(1, 0), v2(0, 1), v3(1, 1);
double t1, t2, s1, s2, w;
Polygon poly;

bool in(const Point& p)
{ return p.x >= t1 && p.x <= t2 && p.y >= s1 && p.y <= s2; }

void check(Point p)
{
    if(in(p) && fabs(p.x-p.y) <= w)//该点要在矩形内两平行线之间
        poly.push_back(Point(p.x, p.y));
}

int main()
{
    //freopen("in.txt", "r", stdin);

    int T;
    scanf("%d", &T);
    for(int kase = 1; kase <= T; kase++)
    {
        poly.clear();
        scanf("%lf%lf%lf%lf%lf", &t1, &t2, &s1, &s2, &w);
        Point p(t1, s2), p1(0, w), p2(0, -w);
        check(p);
        p = Intersection(Point(t1, 0), v2, p1, v3); check(p);
        p = Intersection(Point(t1, 0), v2, p2, v3); check(p);
        p = Point(t1, s1); check(p);
        p = Intersection(Point(0, s1), v1, p1, v3); check(p);
        p = Intersection(Point(0, s1), v1, p2, v3); check(p);
        p = Point(t2, s1); check(p);
        p = Intersection(Point(t2, 0), v2, p2, v3); check(p);
        p = Intersection(Point(t2, 0), v2, p1, v3); check(p);
        p = Point(t2, s2); check(p);
        p = Intersection(Point(0, s2), v1, p2, v3); check(p);
        p = Intersection(Point(0, s2), v1, p1, v3); check(p);
        //for(int i = 0; i < poly.size(); i++) printf("%.3f %.3f
", poly[i].x, poly[i].y);
        double A = PolygonArea(poly);
        double B = (t2-t1) * (s2-s1);
        double ans = A / B;
        printf("Case #%d: %.8f
", kase, ans);
    }

    return 0;
}