题意：知道你和朋友的到达时间，两人最多等W范围分钟，问两人会面的概率
题解：概率空间是一个平面上的矩形，计算面积即可，这里用到了计算几何
#include#include #include #include #include const int maxn = 555;const int maxisn = 10;const double eps = 1e-8;const double pi = acos(-1.0);int dcmp(double x){ if(x > eps) return 1; return x < -eps ? -1 : 0;}inline double min(double a, double b){ return a < b ? a : b;}inline double max(double a, double b){ return a > b ? a : b;}inline double Sqr(double x){ return x * x;}struct Point{ double x, y; Point(){x = y = 0;} Point(double a, double b) {x = a, y = b;} inline Point operator-(const Point &b)const { return Point(x - b.x, y - b.y);} inline Point operator+(const Point &b)const { return Point(x + b.x, y + b.y);} inline double dot(const Point &b)const { return x * b.x + y * b.y;} inline double cross(const Point &b, const Point &c)const { return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);}};Point LineCross(const Point &a, const Point &b, const Point &c, const Point &d){ double u = a.cross(b, c), v = b.cross(a, d); return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v));}double PolygonArea(Point p[], int n){ if(n < 3) return 0.0; double s = p[0].y * (p[n - 1].x - p[1].x); p[n] = p[0]; for(int i = 1; i < n; ++ i) s += p[i].y * (p[i - 1].x - p[i + 1].x); return fabs(s * 0.5);}double CPIA(Point a[], Point b[], int na, int nb)//ConvexPolygonIntersectArea{ Point p[maxisn], tmp[maxisn]; int i, j, tn, sflag, eflag; a[na] = a[0], b[nb] = b[0]; memcpy(p, b, sizeof(Point) * (nb + 1)); for(i = 0; i < na && nb > 2; ++ i) { sflag = dcmp(a[i].cross(a[i + 1], p[0])); for(j = tn = 0; j < nb; ++ j, sflag = eflag) { if(sflag >= 0) tmp[tn ++] = p[j]; eflag = dcmp(a[i].cross(a[i + 1], p[j + 1])); if((sflag ^ eflag) == -2) tmp[tn ++] = LineCross(a[i], a[i + 1], p[j], p[j + 1]); } memcpy(p, tmp, sizeof(Point) * tn); nb = tn, p[nb] = p[0]; } if(nb < 3) return 0.0; return PolygonArea(p, nb);}double SPIA(Point a[], Point b[], int na, int nb)//SimplePolygonIntersectArea{ int i, j; Point t1[4], t2[4]; double res = 0, if_clock_t1, if_clock_t2; a[na] = t1[0] = a[0], b[nb] = t2[0] = b[0]; for(i = 2; i < na; ++ i) { t1[1] = a[i - 1], t1[2] = a[i]; if_clock_t1 = dcmp(t1[0].cross(t1[1], t1[2])); if(if_clock_t1 < 0) std::swap(t1[1], t1[2]); for(j = 2; j < nb; ++ j) { t2[1] = b[j - 1], t2[2] = b[j]; if_clock_t2 = dcmp(t2[0].cross(t2[1], t2[2])); if(if_clock_t2 < 0) std::swap(t2[1], t2[2]); res += CPIA(t1, t2, 3, 3) * if_clock_t1 * if_clock_t2; } } return res;}Point p1[maxn], p2[maxn];int n1, n2;int main(){ double t1,t2,s2,s1,w; int T; scanf("%d", &T); for(int i=1;i<=T;i++){ scanf("%lf%lf%lf%lf%lf", &t1, &t2, &s1, &s2, &w); p1[0].x = t1;p1[0].y = s1; p1[1].x = t1;p1[1].y = s2; p1[2].x = t2;p1[2].y = s2; p1[3].x = t2;p1[3].y = s1; p2[0].x = t1;p2[0].y = t1-w; p2[1].x = t1;p2[1].y = t1+w; p2[2].x = t2;p2[2].y = t2+w; p2[3].x = t2;p2[3].y = t2-w; printf("Case #%d: %.8f\n", i, (SPIA(p1, p2, 4, 4) + eps)/((s2-s1)*(t2-t1))); } return 0;}