Cdoj 27 Baseball Defense (baseball) Problem Solving report

Topic Link HTTP://ACM.UESTC.EDU.CN/#/PROBLEM/SHOW/27

This is a math problem, compared to the pit, it took me a few days

Formula pushed for a long time, because it looks like trouble, so put a few days do not bother to do ... After all summer vacation

The variable name given by the title is too sore, I'm making a new pact here.

The coordinates of the defender are \ ((x_0, y_0) \), Speed \ (v_0\)
The speed of rolling the earth is \ (v\), along the x-axis speed is \ (v_x\), y-axis sub-speed is \ (v_y\)
The landing point of the flying Ball is \ ((x, y) \), flight time \ (t\)

The difficulty lies in rolling the Earth, as long as an infield meets the ball or reaches the trajectory ahead of time.

I didn't ask for a bump in the infield, I thought it was in the infield, rolling the earth.

In this case, the subject becomes a two-time function in \ (\left[0, \frac{r}{\sqrt{x^2+y^2}}\right]\) There is no solution to the situation

If not in the infield is much simpler, only requires a two function in the \ ([0, +\infty) \) There is no solution can

The first flying ball, very simple, calculate the \ (\sqrt{(x-x_0) ^2+ (Y-Y_0) ^2}\) and \ (vt\) relationship can be

This is a simple relationship, I don't think I need an explanation.

Then to roll the Earth, to make the ball accessible, it requires the infield defender to run to the ball time \ (t\) meet the following conditions \[\sqrt{(V_XT-X_0) ^2+ (v_yt-y_0) ^2}\le v_0t\]

This is a two-time inequality about T, we combine both sides of the square and move the items together, and because \ (v_x^2+v_y^2=v^2\), we have: \[(v^2-v_0^2) t^2-2 (v_xx_0+v_yy_0) t+ (x_0^2+y_0^2) \le0\]

Very beautiful inequality, as long as the inequality in \ (t\in [0,\infty) \) has a solution can

When \ (V\ge v_0\), this is obviously a solution

When \ (v < v_0\), the function \ (f (t) = (v^2-v_0^2) t^2-2 (v_xx_0+v_yy_0) t+ (x_0^2+y_0^2) \) has a symmetric axis of \ (t_0=\frac{v_xx_0+v_yy_0}{v^2-v_ 0^2} > 0\)

So just need

\[
\begin{align*}
\delta&=4 (V_XX_0+V_YY_0) ^2-4 (v^2-v_0^2) (x_0^2+y_0^2) \ \
&=4 (v_x^2x_0^2+v_y^2y_0^2+2v_xv_yx_0y_0-v^2x_0^2-v^2y_0^2+v_0^2x_0^2+v_0^2y_0^2) \ \
&=4[v_0^2 (x_0^2+y_0^2)-v_y^2x_0^2+2v_xv_yx_0y_0-v_x^2y_0^2]\\
&=4[v_0^2 (x_0^2+y_0^2)-(V_YX_0-V_XY_0) ^2]\geq0
\end{align*}
\]

i.e. \ (v_0^2 (x_0^2+y_0^2) \geq (v_yx_0-v_xy_0) ^2\)

So in summary, as long as an infield defender satisfies \ (V\ge v_0\) or \ (v_0^2 (x_0^2+y_0^2) \geq (V_YX_0-V_XY_0) ^2\) The strike is out

So hit the code as follows

#include <cstdio> #include <cmath>using namespace std; #define EPS (1e-8) int N, M;int x[10], y[10], V[10];char b All;double XX, YY, T;int H, L, vv;double VX, Vy;bool safe;inline double Sqr (double x) {return x * x;}  int main () {while (scanf ("%d%d", &n, &m) && N && M) {for (int i = 0; i < N; ++i) scanf ("%d%d%d",  &x[i], &y[i], &v[i]); while (m--) {safe = true;while (Ball = GetChar ())! = ' F ' && ball! = ' G '); switch (ball) {case ' G ': scanf ("%d/%d%d", &h, &l, &AMP;VV), VX = VV * L/SQRT (SQR (h) + SQR (l)), VY = vv * H/SQRT (SQR (h) + SQR (l) ); for (int i = 0; i < N; ++i) if (Sqr (X[i]) + SQR (Y[i]) < () safe &= VV > V[i] && (SQR (x[i)) + SQR (y[ I])) * SQR (V[i]) < SQR (X[i] * vy-y[i] * VX); Break;case ' F ': scanf ("%lf%lf%lf", &xx, &yy, &t); for (int i = 0; i < N; ++i) Safe &= sqr (Xx-x[i]) + SQR (Yy-y[i]) > Sqr (v[i] * T); if (safe) puts ("safe"); Else puts ("out");}} return 0;}

Cdoj 27 Baseball Defense (baseball) Problem solving report