poj2074 Line of Sight

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <queue>
 5 
 6 using namespace std;
 7 const int maxn = 1e6;
 8 
 9 double Hx1, Hx2, Hy, Px1, Px2, Py;
10 struct Point{
11     double x1, x2;
12     Point(double x1 = 0, double x2 = 0) : x1(x1), x2(x2) {}
13     bool operator < (const Point& rhs) const{
14         return x1 < rhs.x1 || (x1 == rhs.x1 && x2 < rhs.x2);
15     }
16     Point operator - (const Point& rhs) const{
17         return Point(x1 - rhs.x1, x2 - rhs.x2);
18     }
19 };
20 Point s[maxn];
21 int n, k;
22 double x1, x2, y;
23 
24 double Cross(Point a, Point b)    { return a.x1 * b.x2 - b.x1 * a.x2; }
25 
26 double getPosR(double x, double y){
27     Point vect1 = Point(Hx1, Hy) - Point(x, y);
28     Point vect2 = Point(Px2, Py) - Point(Hx1, Hy);
29     if(Cross(vect1, vect2) >= 0) return Px2;
30     Point vect3 = Point(Px1, Py) - Point(Hx1, Hy);
31     if(Cross(vect1, vect3) <= 0) return Px1;
32     return (x - Hx1) * (Py - Hy) / (y - Hy) + Hx1;
33 }
34 
35 double getPosL(double x, double y){
36     Point vect1 = Point(Hx2, Hy) - Point(x, y);
37     Point vect2 = Point(Px2, Py) - Point(Hx2, Hy);
38     if(Cross(vect1, vect2) >= 0) return Px2;
39     Point vect3 = Point(Px1, Py) - Point(Hx2, Hy);
40     if(Cross(vect1, vect3) <= 0) return Px1;
41     return (x - Hx2) * (Py - Hy) / (y - Hy) + Hx2;
42 }
43 
44 int main(){
45     freopen("in.txt", "r", stdin);
46     while(~scanf("%lf%lf%lf", &Hx1, &Hx2, &Hy) && (Hx1 + Hx2 + Hy)){
47         scanf("%lf%lf%lf", &Px1, &Px2, &Py);
48         scanf("%d", &n);
49         k = 0;
50         for(int i = 0; i < n; i++){
51             scanf("%lf%lf%lf", &x1, &x2, &y);
52             if(y >= Hy || y <= Py) continue;
53             double tx1 = getPosL(x1, y);
54             double tx2 = getPosR(x2, y);
55             if(tx1 != tx2) s[k++] = Point(tx1, tx2);
56         }
57         if(!k){
58             printf("%.2f\n", Px2 - Px1);
59             continue;
60         }
61         sort(s, s + k);
62         double ans = s[0].x1 - Px1;
63         for(int i = 0; i < k; i++){
64             double rear = s[i].x2;
65             int j = i;
66             while(i + 1 < k && s[i + 1].x2 <= rear) ++i;
67             if(i == k - 1){
68                 ans = max(ans, Px2 - rear);
69                 break;
70             }
71             double front = s[i + 1].x1;
72             if(front > rear) ans = max(ans, front - rear);
73         }
74         if(!ans) puts("No View");
75         else printf("%.2f\n", ans);
76     }
77     return 0;
78 }