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，=_=||

 1 #include <cstdio>

 2 #include <vector>

 3 #include <cmath>

 4 using namespace std;

 5 

 6 struct Point

 7 {

 8     double x, y;

 9     Point(double x=0, double y=0):x(x), y(y) {}

10 };

11 typedef Point Vector;

12 typedef vector<Point> Polygon;

13 

14 Point operator + (const Point& A, const Point& B)

15 { return  Point(A.x+B.x, A.y+B.y); }

16 

17 Vector operator - (const Point& A, const Point& B)

18 { return  Point(A.x-B.x, A.y-B.y); }

19 

20 Vector operator * (const Vector& A, double p)

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

22 

23 double Cross(const Vector& A, const Vector& B)

24 { return A.x * B.y - A.y * B.x; }

25 

26 double PolygonArea(const Polygon& p)

27 {//求多边形面积

28      int n = p.size();

29      double area = 0;

30      for(int i = 1; i < n-1; i++)

31         area += Cross(p[i]-p[0], p[i+1]-p[0]);

32     return area/2;

33 }

34 

35 Point Intersection(Point P, Vector v, Point Q, Vector w)

36 {//求两直线交点

37     Vector u = P-Q;

38     double t = Cross(w, u) / Cross(v, w);

39     return P+v*t;

40 }

41 

42 const double sqrt2 = sqrt(2.0);

43 const Vector v1(1, 0), v2(0, 1), v3(1, 1);

44 double t1, t2, s1, s2, w;

45 Polygon poly;

46 

47 bool in(const Point& p)

48 { return p.x >= t1 && p.x <= t2 && p.y >= s1 && p.y <= s2; }

49 

50 void check(Point p)

51 {

52     if(in(p) && fabs(p.x-p.y) <= w)//该点要在矩形内两平行线之间

53         poly.push_back(Point(p.x, p.y));

54 }

55 

56 int main()

57 {

58     //freopen("in.txt", "r", stdin);

59 

60     int T;

61     scanf("%d", &T);

62     for(int kase = 1; kase <= T; kase++)

63     {

64         poly.clear();

65         scanf("%lf%lf%lf%lf%lf", &t1, &t2, &s1, &s2, &w);

66         Point p(t1, s2), p1(0, w), p2(0, -w);

67         check(p);

68         p = Intersection(Point(t1, 0), v2, p1, v3); check(p);

69         p = Intersection(Point(t1, 0), v2, p2, v3); check(p);

70         p = Point(t1, s1); check(p);

71         p = Intersection(Point(0, s1), v1, p1, v3); check(p);

72         p = Intersection(Point(0, s1), v1, p2, v3); check(p);

73         p = Point(t2, s1); check(p);

74         p = Intersection(Point(t2, 0), v2, p2, v3); check(p);

75         p = Intersection(Point(t2, 0), v2, p1, v3); check(p);

76         p = Point(t2, s2); check(p);

77         p = Intersection(Point(0, s2), v1, p2, v3); check(p);

78         p = Intersection(Point(0, s2), v1, p1, v3); check(p);

79         //for(int i = 0; i < poly.size(); i++) printf("%.3f %.3f\n", poly[i].x, poly[i].y);

80         double A = PolygonArea(poly);

81         double B = (t2-t1) * (s2-s1);

82         double ans = A / B;

83         printf("Case #%d: %.8f\n", kase, ans);

84     }

85 

86     return 0;

87 }

代码君