poj 1039 Pipe (Geometry)
理解错题意一个晚上。_(:з」∠)_
题意很容易看懂，就是要求你求出从外面射进一根管子的射线，最远可以射到哪里。
正解的做法是，选择上点和下点各一个，然后对于每个折点位置竖直位置判断经过的点是否在管中。如果是，就继续找，如果不在管中，这时射线必然已经穿过管出去了，这时就要找射线和管上下壁的交点。
代码如下：
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #include <cmath> 7 8 using namespace std; 9 10 const double EPS = 1e-6; 11 const int N = 33; 12 inline int sgn(double x) { return (x > EPS) - (x < -EPS);} 13 struct Point { 14 double x, y; 15 Point() {} 16 Point(double x, double y) : x(x), y(y) {} 17 bool operator < (Point a) const { return sgn(x - a.x) < 0 || sgn(x - a.x) == 0 && y < a.y;} 18 bool operator == (Point a) const { return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;} 19 Point operator + (Point a) { return Point(x + a.x, y + a.y);} 20 Point operator - (Point a) { return Point(x - a.x, y - a.y);} 21 Point operator * (double p) { return Point(x * p, y * p);} 22 Point operator / (double p) { return Point(x / p, y / p);} 23 } ; 24 typedef Point Vec; 25 inline double dot(Vec a, Vec b) { return a.x * b.x + a.y * b.y;} 26 inline double cross(Vec a, Vec b) { return a.x * b.y - a.y * b.x;} 27 inline double veclen(Vec x) { return sqrt(dot(x, x));} 28 inline Point vecunit(Vec x) { return x / veclen(x);} 29 inline Point normal(Vec x) { return Point(-x.y, x.x) / veclen(x);} 30 31 struct Line { 32 Point s, t; 33 Line() {} 34 Line(Point s, Point t) : s(s), t(t) {} 35 Vec vec() { return t - s;} 36 Point point(double p) { return s + vec() * p;} 37 } ; 38 39 40 Point up[N], dw[N]; 41 Line ul[N], dl[N]; 42 43 inline Point llint(Point P, Vec u, Point Q, Vec v) { return P + u * (cross(v, P - Q) / cross(u, v));} 44 bool tstcross(Point a, Point b, Point c, Point d) { return sgn(cross(a - c, b - c)) * sgn(cross(a - d, b - d)) > 0;} 45 46 double cal(Point s, Point t, int n) { 47 Line tl = Line(s, t); 48 if (tstcross(tl.s, tl.t, up[0], dw[0])) return -1e100; 49 for (int i = 0; i < n - 1; i++) { 50 if (tstcross(tl.s, tl.t, up[i + 1], dw[i + 1])) { 51 double ret = -1e100; 52 if (!tstcross(tl.s, tl.t, ul[i].s, ul[i].t)) { 53 Point tp = llint(tl.s, tl.vec(), ul[i].s, ul[i].vec()); 54 ret = max(ret, tp.x); 55 } 56 if (!tstcross(tl.s, tl.t, dl[i].s, dl[i].t)) { 57 Point tp = llint(tl.s, tl.vec(), dl[i].s, dl[i].vec()); 58 ret = max(ret, tp.x); 59 } 60 return ret; 61 } 62 } 63 return 1e100; 64 } 65 66 void work(int n) { 67 for (int i = 0; i < n - 1; i++) { 68 ul[i] = Line(up[i], up[i + 1]); 69 dl[i] = Line(dw[i], dw[i + 1]); 70 } 71 double ans = -1e100; 72 for (int i = 0; i < n; i++) { 73 for (int j = i + 1; j < n; j++) { 74 if (ans >= 1e99) break; 75 ans = max(ans, cal(up[i], dw[j], n)); 76 ans = max(ans, cal(dw[i], up[j], n)); 77 } 78 } 79 if (ans >= 1e99) puts("Through all the pipe."); 80 else printf("%.2f\n", ans); 81 } 82 83 int main() { 84 // freopen("in", "r", stdin); 85 int n; 86 while (~scanf("%d", &n) && n) { 87 for (int i = 0; i < n; i++) { 88 scanf("%lf%lf", &up[i].x, &up[i].y); 89 dw[i] = Point(up[i].x, up[i].y - 1.0); 90 } 91 work(n); 92 } 93 return 0; 94 }
——written by Lyon
