Poj1039 -- Pipe (Computational ry)

Poj1039 -- Pipe (Computational ry)

 

A pipe has n fold points. The coordinates (x1, y1) (x2, y2) of the n points on the wall of the pipe are given.

If you want to make the light the farthest, in addition to the possibility of the whole process, light will surely be at the intersection of a pipe wall and will be at the intersection of two nodes in the pipe. Otherwise, you can adjust the light angle, let the image go further, enumerate the positions of the two points, and then judge the intersection position one by one.

 

# Include
 
  
# Include
  
   
# Include
   
    
# Include using namespace std; # define eqs 1e-9/* intersection of two straight lines, seeking focus ax + by + c = 0; ux + vy + w = 0; a/u! = B/v prevents parallel intersections: (B * w-v * c)/(a * v-u * B), (c * u-a * w) /(a * v-u * B) */struct point {double x, y;} p [30]; int n; double a [4] [4] = {0, 0, 0}, {0,-, 0}, {0, 0,-1}, {0, -1 }}; double solve (point u, point v) {double a = (u. y-v.y)/(u. x-v.x), B = u. y-a * u. x; double x, y, ans = 0.0; int flag = 0, I; y = a * p [0]. x + B; if (y-p [0]. y> = eqs | p [0]. y-1.0-y> = eqs) return ans; for (I = 1; I <n; I ++) {y = A * p [I]. x + B; if (y-p [I]. y> eqs) {flag = 1; break;} else if (p [I]. y-1.0-y> eqs) {flag = 2; break;} else ans + = (p [I]. x-p [I-1]. x);} if (! Flag) return ans; u = p [I-1], v = p [I]; if (flag = 2) u. y-= 1.0, v. y-= 1.0; double k = (u. y-v.y)/(u. x-v.x), c = u. y-k * u. x; x = (B-c)/(k-a); ans + = (x-p [I-1]. x); return ans;} int main () {int I, j, k; double max1, temp; point u, v; while (scanf (% d, & n) & n) {max1 = 0.0; for (I = 0; I <n; I ++) scanf (% lf, & p [I]. x, & p [I]. y); for (I = 0; I <n; I ++) for (j = I + 1; j <n; j ++) for (k = 0; k <4; k ++) {u. x = p [I]. x + a [k] [0]; u. y = p [I]. y + a [k] [1]; v. x = p [j]. x + a [k] [2]; v. y = p [j]. y + a [k] [3]; temp = solve (u, v); if (temp> max1) max1 = temp ;}if (fabs (p [n-1]. x-p [0]. x-max1) <eqs) printf (Through all the pipe .); else printf (%. 2lf, p [0]. x + max1) ;}return 0 ;}