POJ 1556 - The Doors 线段相交不含端点

POJ 1556 - The Doors
题意：
    在 10x10 的空间里有很多垂直的墙，不能穿墙，问你从(0,5) 到 (10,5)的最短距离是多少.
    
分析：
    
    要么直达，要么一定是墙的边缘点之间以及起始点、终点的连线.
    
    所以先枚举墙上每一点到其他点的直线可达距离，就是要判定该线段是否与墙相交(不含端点).
    
    然后最短路.

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 using namespace std;
 const double INF = 1e10;
 const double eps = 1e-;
 int dcmp(double x)
 {
     return fabs(x) < eps ?  : (x <  ? - : );
 }
 struct Point
 {
     double x,y;
     Point(double x1 = ,double y1 = ) : x(x1), y(y1) {}
 };
 Point operator - (Point a,Point b)
 {
     return Point(a.x - b.x, a.y - b.y);
 }
 double Det(Point a,Point b)
 {
     return a.x * b.y - a.y * b.x;
 }
 double Dot(Point a,Point b)
 {
     return a.x * b.x + a.y * b.y;
 }
 double Length(Point a)
 {
     return sqrt(Dot(a, a));
 }
 bool OnSegment(Point p, Point a1, Point a2)
 {
     return dcmp(Det(a1 - p, a2 - p)) ==  && dcmp(Dot(a1 - p, a2 - p) ) <= ;
 }
 struct Line
 {
     Point s,e;
     Line() {}
     Line(Point s1, Point e1) : s(s1), e(e1) {}
 };
 bool SegCross(Point a1, Point a2, Point b1, Point b2)
 {
     double c1 = Det(a2 - a1, b1 - a1);
     double c2 = Det(a2 - a1, b2 - a1);
     double c3 = Det(b2 - b1, a1 - b1);
     double c4 = Det(b2 - b1, a2 - b1);
     if(dcmp(c1) * dcmp(c2) <  && dcmp(c3) * dcmp(c4) < ) return ;
     else return ;
 }

 int n;
 Line l[];
 Point p[];
 double map[][];
 int main()
 {
     while (~scanf("%d", &n) && n != -)
     {
         p[] = Point(, );
         p[] = Point(, );
         int t = , m = ;
         for (int i = ; i <= n; i++)
         {
             double x; Point p1,p2,p3,p4;
             scanf("%lf%lf%lf%lf%lf",&x, &p1.y, &p2.y, &p3.y, &p4.y);
             p1.x = p2.x = p3.x = p4.x = x;
             l[m++] = Line(Point(x,),p1);
             l[m++] = Line(p2, p3);
             l[m++] = Line(p4, Point(x,));
             p[t++] = p1; p[t++] = p2; p[t++] = p3; p[t++] = p4;
         }
         for (int i = ; i < t; i++)
             for (int j = ; j < t; j++)
                 map[i][j] = INF;
         for (int i = ; i < t; i++)
         {
             for (int j = i+; j < t; j++)
             {
                 if (p[i].x == p[j].x) continue;
                 bool flag = ;
                 for (int k = ; k < m; k++)//枚举墙
                 {
                     if(SegCross(p[i], p[j], l[k].e, l[k].s))
                     {
                         flag = ; break;
                     }
                 }
                 if (flag)
                 {
                     map[i][j] = map[j][i] = Length(p[i]-p[j]);
                 }

             }
         }
         for(int i = ; i < t; i++) map[i][i] = ;
         for (int k = ; k < t; k++)
             for (int i = ; i < t;i++)
                 for (int j = ; j < t; j++)
                     map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
         printf("%.2f\n",map[][]);
     }
 }