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=, double y=):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 = ;
      for(int i = ; i < n-; i++)
         area += Cross(p[i]-p[], p[i+]-p[]);
     return area/;
 }

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

     return ;
 }

代码君