POJ 1556 - The Doors 线段相交不含端点

最新推荐文章于 2021-05-01 21:09:13 发布

weixin_33862993

最新推荐文章于 2021-05-01 21:09:13 发布

阅读量128

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/nicetomeetu/p/5762959.html

本文提供了一种解决POJ1556-TheDoors问题的方法，该问题要求找到在一个充满垂直障碍物的空间中两点间的最短路径。通过枚举所有可能的路径并使用几何算法判断路径的有效性，最终利用Floyd算法找到了最优解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <cstring>
  5 using namespace std;
  6 const double INF = 1e10;
  7 const double eps = 1e-10;
  8 int dcmp(double x)
  9 {
 10     return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1);
 11 }
 12 struct Point
 13 {
 14     double x,y;
 15     Point(double x1 = 0,double y1 = 0) : x(x1), y(y1) {}
 16 };
 17 Point operator - (Point a,Point b)
 18 {
 19     return Point(a.x - b.x, a.y - b.y);
 20 }
 21 double Det(Point a,Point b)
 22 {
 23     return a.x * b.y - a.y * b.x; 
 24 }
 25 double Dot(Point a,Point b)
 26 {
 27     return a.x * b.x + a.y * b.y;
 28 }
 29 double Length(Point a)
 30 {
 31     return sqrt(Dot(a, a));
 32 }
 33 bool OnSegment(Point p, Point a1, Point a2)
 34 {
 35     return dcmp(Det(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p) ) <= 0;
 36 }
 37 struct Line
 38 {
 39     Point s,e;
 40     Line() {}
 41     Line(Point s1, Point e1) : s(s1), e(e1) {}
 42 };
 43 bool SegCross(Point a1, Point a2, Point b1, Point b2)
 44 {
 45     double c1 = Det(a2 - a1, b1 - a1);
 46     double c2 = Det(a2 - a1, b2 - a1);
 47     double c3 = Det(b2 - b1, a1 - b1);
 48     double c4 = Det(b2 - b1, a2 - b1);
 49     if(dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0) return 1;
 50     else return 0;
 51 }
 52 
 53 int n;
 54 Line l[100];
 55 Point p[100];
 56 double map[100][100];
 57 int main()
 58 {
 59     while (~scanf("%d", &n) && n != -1)
 60     {
 61         p[0] = Point(0, 5);
 62         p[1] = Point(10, 5);
 63         int t = 2, m = 0;
 64         for (int i = 1; i <= n; i++)
 65         {
 66             double x; Point p1,p2,p3,p4;
 67             scanf("%lf%lf%lf%lf%lf",&x, &p1.y, &p2.y, &p3.y, &p4.y);
 68             p1.x = p2.x = p3.x = p4.x = x;
 69             l[m++] = Line(Point(x,0),p1);
 70             l[m++] = Line(p2, p3);
 71             l[m++] = Line(p4, Point(x,10));
 72             p[t++] = p1; p[t++] = p2; p[t++] = p3; p[t++] = p4;
 73         }
 74         for (int i = 0; i < t; i++)
 75             for (int j = 0; j < t; j++)
 76                 map[i][j] = INF;
 77         for (int i = 0; i < t; i++)
 78         {
 79             for (int j = i+1; j < t; j++)
 80             {
 81                 if (p[i].x == p[j].x) continue;
 82                 bool flag = 1; 
 83                 for (int k = 0; k < m; k++)//枚举墙
 84                 {
 85                     if(SegCross(p[i], p[j], l[k].e, l[k].s)) 
 86                     {
 87                         flag = 0; break;
 88                     }
 89                 } 
 90                 if (flag)
 91                 {
 92                     map[i][j] = map[j][i] = Length(p[i]-p[j]);
 93                 }
 94                     
 95             }
 96         }
 97         for(int i = 0; i < t; i++) map[i][i] = 0;
 98         for (int k = 0; k < t; k++)
 99             for (int i = 0; i < t;i++)
100                 for (int j = 0; j < t; j++)
101                     map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
102         printf("%.2f\n",map[0][1]);
103     }
104 }