HDU5572 2015上海现场赛A题1001
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 #include<map> 6 #include<set> 7 #include<cmath> 8 using namespace std; 9 typedef long long LL; 10 const double eps = 1e-8; 11 int sgn(double x) { 12 if(fabs(x) < eps) return 0; 13 if(x < 0) return -1; 14 else return 1; 15 } 16 struct Point { 17 double x, y; 18 Point(double x = 0, double y = 0) : x(x), y(y) {} 19 Point operator - (const Point& t) const { 20 return Point(x - t.x, y - t.y); 21 } 22 double operator ^ (const Point& t) const { 23 return x * t.y - y * t.x; 24 } 25 double operator * (const Point& t) const { 26 return x * t.x + y * t.y; 27 } 28 Point operator * (const double& k) const { 29 return Point(x * k, y * k); 30 } 31 Point operator / (const double& k) const { 32 return Point(x / k, y / k); 33 } 34 Point operator + (const Point& t) const { 35 return Point(x + t.x, y + t.y); 36 } 37 double distance(Point p) { 38 return hypot(x - p.x, y - p.y); 39 } 40 double len() { 41 return hypot(x, y); 42 } 43 double len2() { 44 return x * x + y * y; 45 } 46 Point trunc(double r) { 47 double l = len(); 48 if(!sgn(l)) return *this; 49 r /= l; 50 return Point(x * r, y * r); 51 } 52 }; 53 struct Line { 54 Point s, e; 55 Line() {} 56 Line(Point s, Point e) { 57 this->s = s; this->e = e; 58 } 59 double length() { 60 return sqrt((s.x - e.x) * (s.x - e.x) + (s.y - e.y) * (s.y - e.y)); 61 } 62 int relation(Point p) { 63 int c = sgn((p - s) ^ (e - s)); 64 if(c < 0) return 1; 65 else if(c > 0) return 2; 66 else return 3; 67 } 68 double dispointtoline(Point p) { 69 return fabs((p - s) ^ (e - s)) / length(); 70 } 71 double dispointtoseg(Point p) { 72 if(sgn((p - s) * (e - s)) < 0 || sgn((p - e) * (s - e)) < 0) 73 return min(p.distance(s), p.distance(e)); 74 return dispointtoline(p); 75 } 76 Point lineprog(Point p) { 77 return s + (((e - s) * ((e - s) * (p - s))) / ((e - s).len2())); 78 } 79 Point symmetrypoint(Point p) { 80 Point q = lineprog(p); 81 return Point(2 * q.x - p.x, 2 * q.y - p.y); 82 } 83 }; 84 struct Circle { 85 Point p; 86 double r; 87 Circle(double x, double y, double r) { 88 p = Point(x, y); 89 this->r = r; 90 } 91 int relationline(Line v) { 92 double dst = v.dispointtoline(p); 93 if(sgn(dst - r) < 0) return 2; 94 else if(sgn(dst - r) == 0) return 1; 95 else return 0; 96 } 97 int relationseg(Line v) { 98 double dst = v.dispointtoseg(p); 99 if(sgn(dst - r) < 0) return 2; 100 else if(sgn(dst - r) == 0) return 1; 101 else return 0; 102 } 103 int pointcrossline(Line v, Point& p1, Point& p2) { 104 if(!(*this).relationline(v)) return 0; 105 Point a = v.lineprog(p); 106 double d = v.dispointtoline(p); 107 d = sqrt(r * r - d * d); 108 if(sgn(d) == 0) { 109 p1 = a; p2 = a; 110 return 1; 111 } 112 p1 = a + (v.e - v.s).trunc(d); 113 p2 = a - (v.e - v.s).trunc(d); 114 return 2; 115 } 116 }; 117 int main() { 118 int T; scanf("%d", &T); 119 int ca(1); 120 while(T--) { 121 int ox, oy, r; 122 int ax, ay, vx, vy; 123 int bx, by; 124 scanf("%d%d%d", &ox, &oy, &r); 125 scanf("%d%d%d%d", &ax, &ay, &vx, &vy); 126 scanf("%d%d", &bx, &by); 127 Line l(Point(ax, ay), Point(ax + vx, ay + vy)); 128 Circle c(ox, oy, r); 129 Point d(bx, by); 130 Line t(Point(ax, ay), Point(bx, by)); 131 if(l.relation(d) == 3 && c.relationseg(t) <= 1 && sgn(Point(bx - ax, by - ay) ^ Point(vx, vy)) == 0) printf("Case #%d: Yes\n", ca++); 132 else { 133 Point a, b; 134 if(c.pointcrossline(l, a, b) != 2) printf("Case #%d: No\n", ca++); 135 else { 136 Point cut; 137 if(a.distance(Point(ax, ay)) > b.distance(Point(ax, ay))) cut = b; 138 else cut = a; 139 Line mid(cut, c.p); 140 Point en = mid.symmetrypoint(Point(ax, ay)); 141 Line aft(cut, en); 142 if(aft.e.distance(d) > aft.s.distance(d)) swap(aft.s, aft.e); 143 if(sgn((aft.e - aft.s) ^ Point(d.x - cut.x, d.y - cut.y)) == 0) printf("Case #%d: Yes\n", ca++); 144 else printf("Case #%d: No\n", ca++); 145 } 146 } 147 } 148 return 0; 149 }
醉了，，现场读错题了，分反弹和不反弹两种情况讨论就好了，加点特判
