poj 1039 Pipe (Geometry)

1039 -- Pipe

　　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：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <cmath>

 using namespace std;

 const double EPS = 1e-;
 const int N = ;
 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
 struct Point {
     double x, y;
     Point() {}
     Point(double x, double y) : x(x), y(y) {}
     bool operator < (Point a) const { return sgn(x - a.x) <  || sgn(x - a.x) ==  && y < a.y;}
     bool operator == (Point a) const { return sgn(x - a.x) ==  && sgn(y - a.y) == ;}
     Point operator + (Point a) { return Point(x + a.x, y + a.y);}
     Point operator - (Point a) { return Point(x - a.x, y - a.y);}
     Point operator * (double p) { return Point(x * p, y * p);}
     Point operator / (double p) { return Point(x / p, y / p);}
 } ;
 typedef Point Vec;
 inline double dot(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}
 inline double cross(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}
 inline double veclen(Vec x) { return sqrt(dot(x, x));}
 inline Point vecunit(Vec x) { return x / veclen(x);}
 inline Point normal(Vec x) { return Point(-x.y, x.x) / veclen(x);}

 struct Line {
     Point s, t;
     Line() {}
     Line(Point s, Point t) : s(s), t(t) {}
     Vec vec() { return t - s;}
     Point point(double p) { return s + vec() * p;}
 } ;

 Point up[N], dw[N];
 Line ul[N], dl[N];

 inline Point llint(Point P, Vec u, Point Q, Vec v) { return P + u * (cross(v, P - Q) / cross(u, v));}
 bool tstcross(Point a, Point b, Point c, Point d) { return sgn(cross(a - c, b - c)) * sgn(cross(a - d, b - d)) > ;}

 double cal(Point s, Point t, int n) {
     Line tl = Line(s, t);
     if (tstcross(tl.s, tl.t, up[], dw[])) return -1e100;
     for (int i = ; i < n - ; i++) {
         if (tstcross(tl.s, tl.t, up[i + ], dw[i + ])) {
             double ret = -1e100;
             if (!tstcross(tl.s, tl.t, ul[i].s, ul[i].t)) {
                 Point tp = llint(tl.s, tl.vec(), ul[i].s, ul[i].vec());
                 ret = max(ret, tp.x);
             }
             if (!tstcross(tl.s, tl.t, dl[i].s, dl[i].t)) {
                 Point tp = llint(tl.s, tl.vec(), dl[i].s, dl[i].vec());
                 ret = max(ret, tp.x);
             }
             return ret;
         }
     }
     return 1e100;
 }

 void work(int n) {
     for (int i = ; i < n - ; i++) {
         ul[i] = Line(up[i], up[i + ]);
         dl[i] = Line(dw[i], dw[i + ]);
     }
     double ans = -1e100;
     for (int i = ; i < n; i++) {
         for (int j = i + ; j < n; j++) {
             if (ans >= 1e99) break;
             ans = max(ans, cal(up[i], dw[j], n));
             ans = max(ans, cal(dw[i], up[j], n));
         }
     }
     if (ans >= 1e99) puts("Through all the pipe.");
     else printf("%.2f\n", ans);
 }

 int main() {
 //    freopen("in", "r", stdin);
     int n;
     while (~scanf("%d", &n) && n) {
         for (int i = ; i < n; i++) {
             scanf("%lf%lf", &up[i].x, &up[i].y);
             dw[i] = Point(up[i].x, up[i].y - 1.0);
         }
         work(n);
     }
     return ;
 }

——written by Lyon