Hdu_4170 Supply Mission (Computational ry)

Question:
The aircraft is in the position (x0, y0), the flight speed is v km/h, there are N (0 <N <8) ships are (xi, yi) the velocity vector is (vxi, vyi) km/h, and the coordinate unit is km.
The plane must unload the cargo for one hour on each ship, and finally fly back to the original location (x0, y0). The minimum time is required and the time is output in minutes and seconds.
Ideas:
After thinking about it for a while, I have no idea. I looked at the data range. N is the maximum 8, and the time limit is 10 seconds. It can be completely violent.
Directly arrange the order of arrival to the ship, and take the minimum time. Be careful when calculating the encounter.
There is also a small trick, which is equal to 60 after the second is carried. It's been a long time for me here...
[Cpp]
/*
Program: hdu_4170 & hunnu_000015
Author: BlackAndWhite
*/
# Include <stdio. h>
# Include <math. h>
# Include <algorithm>
# Define eps 1e-6
Using namespace std;
Struct point
{
Int x, y;
} A [10], v [10], s;
Double t, ans;
Int N, sv, I, p [10], T [3], js = 1;
Double crosstime (double x0, double y0, double v0, double x1, double y1, double vx, double vy)
{// It indicates the time when the plane is in (x0, y0) and the speed is v0. The time it takes to meet the velocity vector of the ship (x1, y1) is (vx, vy.
Double a, B, c, t;
A = vx * vx + vy * vy-v0 * v0;
B = 2 * (vx * (x1-x0) + vy * (y1-y0 ));
C = (x1-x0) * (x1-x0) + (y1-y0) * (y1-y0 );
If (a <eps & a>-eps) return-c/B;
If (a <eps) {a =-a; B =-B; c =-c ;}
Return (-B + sqrt (B * B-4 * a * c)/(2 * );
}
Double back (double x0, double y0, double x1, double y1, double sv) // time spent returning to the start point
{
Return sqrt (x1-x0) * (x1-x0) + (y1-y0) * (y1-y0)/sv;
}
Void trs (double ans, int T [3]) // converts decimal form to hour/minute/second
{
T [0] = (int) ans;
Ans-= T [0];
T [1] = (int) (ans * = 60.0); ans-= T [1];
T [2] = (int) (ans x 60.0 + 0.9999999 );
If (T [2] = 60) {T [2] = 0; T [1] ++;} // trick, carry...
If (T [1] = 60) {T [1] = 0; T [0] ++ ;}
}
Int main ()
{
While (scanf ("% d", & N), N)
{
For (I = 0; I <N; I ++) scanf ("% d", & a [I]. x, & a [I]. y, & v [I]. x, & v [I]. y );
Scanf ("% d", & s. x, & s. y, & sv );
For (I = 0; I <= N; I ++) p [I] = I;
Ans = 1e9;
Do
{
T = crosstime (s. x * 1.0, s. y * 1.0, sv * 1.0, a [p [0]. x, a [p [0]. y, v [p [0]. x, v [p [0]. y) + 1.0;
For (I = 1; I <N; I ++) // The current departure moment on the p [I-1] ship is ready, p [I-1] has been traveling t hours, p [I] the same t hour
T + = 1.0 + crosstime (a [p [I-1]. x * 1.0 + t * v [p [I-1]. x, a [p [I-1]. y * 1.0 + t * v [p [I-1]. y, sv * 1.0, a [p [I]. x * 1.0 + t * v [p [I]. x, a [p [I]. y * 1.0 + t * v [p [I]. y, v [p [I]. x * 1.0, v [p [I]. y * 1.0 );
T + = back (a [p [N-1]. x * 1.0 + t * v [p [N-1]. x, a [p [N-1]. y * 1.0 + t * v [p [N-1]. y, s. x, s. y, sv );
If (t <ans) ans = t;
}
While (next_permutation (p, p + N ));
Trs (ans, T );
Printf ("Case % d: % d hour (s) % d minute (s) % d second (s) \ n", js ++, T [0], T [1], T [2]);
}
Return 0;
}