Pipe POJ - 1039 (intersection)

Pipe   POJ - 1039

Topic links: https://vjudge.net/problem/POJ-1039

The meaning of problems: given the coordinates of each point on you tube wall, the lower wall of each corresponding point of the same abscissa, the ordinate of the upper wall ordinate minus one, seeking whether a light entirely through the conduit, if not completely, to obtain the maximum value

Idea: It is a subject line intersects the line segment, if there is a line which has a maximum value, then certainly by translating this line enables rotation through a vertex and a vertex in the upper wall of the lower wall, so that we can through all points, and the vertex line connected into the lower vertices, the determination of the bending line and the other pipe intersect, should not want to pay, then it is completely through the reservation if the maximum value can intersect, each intersection.

 1 // 
 2 // Created by HJYL on 2020/1/14.
 3 //
 4 #include <stdio.h>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <cmath>
 8 #include <vector>
 9 #include <string>
10 #include <cstring>
11 #include <iomanip>
12 using namespace std;
13 const int maxn=100+10;
14 const double eps=1e-8;
15 struct Point{
16     double x,y;
17 };
18 
19 double Cross(Point p1,Point p2,Point p0)
20 {
21     return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
22 }
23 
24 bool zx(Point a,Point b,Point c,Point d)
25 {
26     double tmp=Cross(c,a,b)*Cross(d,a,b);
27     if(fabs(tmp)<eps||tmp<0)
28         return true;
29     else
30         return false;
31 }
32 
33 double getpoint(Point a,Point b,Point c,Point d)
34 {
35     double aa=b.y-a.y;
36     double bb=a.x-b.x;
37     double cc=(b.x-a.x)*a.y-(b.y-a.y)*a.x;
38     double aaa=d.y-c.y;
39     double bbb=c.x-d.x;
40     double ccc=(d.x-c.x)*c.y-(d.y-c.y)*c.x;
41     double x=(ccc*bb-cc*bbb)/(aa*bbb-aaa*bb);
42     double y=(cc*aaa-ccc*aa)/(aa*bbb-aaa*bb);
43     return x;
44 }
45 
46 int main()
47 {
48     int n;
49     bool flag;
50     Point above[maxn],below[maxn];
51 
52     while(~scanf("%d",&n)&&n)
53     {
54         for(int i=0;i<n;i++)
55         {
56             scanf("%lf%lf",&above[i].x,&above[i].y);
57             below[i].x=above[i].x;
58             below[i].y=above[i].y-1;
59         }
60         double head=above[0].x;
61         flag=false;
62         int k;
63         for(int i=0;i<n&&!flag;i++)
64         {
65             for(int j=0;j<n&&!flag;j++)
66             {
67                 if(i==j)
68                     break;
69                 for(k=0;k<n;k++)
70                 {
71                     if(!zx(above[i],below[j],above[k],below[k]))//ij连成的线是否与k管道相交
72                         break;
73                 }
74                 if(k==n)
75                     flag=true;
76                 else if(k>max(i,j))
77                 {
78                     head=max(head,getpoint(above[i],below[j],above[k-1],above[k]));
79                     head=max(head,getpoint(above[i],below[j],below[k-1],below[k]));
80                 }
81             }
82         }
83         if(flag)
84             printf("Through all the pipe.\n");
85         else
86             printf("%.2lf\n",head);
87     }
88     return 0;
89 }