poj 1039 Pipe（几何基础）

 1 #include<cmath>
 2 #include<cstdio> 
 3 #include<algorithm>
 4 using namespace std;
 5 const int N = 22;
 6 const double eps = 1e-8;
 7  
 8 struct Pt{
 9     double x, y;
10 }a[N], b[N];
11 struct Line{ double a, b, c; };
12 int n;
13 double ans;
14 
15 double mult(Pt sp, Pt ep, Pt op){
16     return (sp.x-op.x)*(ep.y-op.y) - (ep.x-op.x)*(sp.y-op.y);
17 }
18 Line getLine(Pt p1, Pt p2){
19     Line ans;
20     ans.a = p1.y - p2.y;
21     ans.b = p2.x - p1.x;
22     ans.c = p1.x*p2.y - p2.x*p1.y;
23     return ans;
24 }
25 
26 bool solve(Pt p1, Pt p2, int e){
27     int i, flag;
28     for(i = 0; i < n-1; i ++) {
29         if(mult(p2, a[i], p1) < -eps || mult(p2, a[i+1], p1) < -eps){
30             flag = 1; break;
31         }
32         if(mult(p2, b[i], p1) > eps || mult(p2, b[i+1], p1) > eps){
33             flag = 2; break;
34         }
35     }
36     if(i == n-1) return true;   //  没有与任何的管道相交，Through all the pipe.
37     if(i < e) return false;     //  光线不合法。
38     Line l1, l2;                //  光线合法，求出射到的最远距离。
39     l1 = getLine(p1, p2);
40     if(flag == 1) l2 = getLine(a[i], a[i+1]);
41     else l2 = getLine(b[i], b[i+1]);
42     ans = max(ans, (l1.b*l2.c-l2.b*l1.c)/(l1.a*l2.b-l2.a*l1.b));
43     return false;
44 }
45 
46 int main(){
47     int i, j;
48     while(scanf("%d", &n) && n){
49         for(i = 0; i < n; i ++){
50             scanf("%lf%lf", &a[i].x, &a[i].y);
51             b[i].x = a[i].x;
52             b[i].y = a[i].y - 1;
53         }
54         ans = -1e9;
55         bool flag = 0;
56         if(n < 3) flag = 1;
57         for(i = 0; i < n; i ++) {
58             for(j = i + 1; j < n; j ++){
59                 flag = solve(a[i], b[j], j);
60                 if(flag) break;
61                 flag = solve(b[i], a[j], j);
62                 if(flag) break;
63             }
64             if(flag) break;
65         }
66         if(flag) puts("Through all the pipe.");
67         else printf("%.2lf\n", ans);
68     }
69     return 0;
70 }