HDU 2857 mirror and light

There is a mirror, and a bunch of light comes in. We know that the mirror coordinates (x1, Y1) (X2, Y2) are two points to determine the mirror surface, the incident point of light and the reflected point of light. Calculate the coordinates of the points on the mirror.

Idea: for example

(1) Straight Line (x1, Y1) (X2, Y2)

(2) obtain the symmetry points (TMP. X, TMP. Y) of a straight line (x1, Y1) (X2, Y2)

(3) evaluate a straight line (TMP. X, TMP. Y) (E. X, E. Y)

(4) Calculate the intersection of (x1, Y1) (X2, Y2) and (TMP. X, TMP. Y) (E. X, E. Y) (x, y)

Formula:

· If two points (x1, Y1) (X2, Y2) on a known straight line are known, the straight line AX + by + c = 0 is obtained.

A = y2-y1; B = x1-x2; C = Y1 * x2-x1 * Y2;

· Known (x0, y0), obtain the symmetric point (x, y) about the straight line AX + by + c = 0)

K =-2 * (A * x0 + B * y0 + C)/(A * A + B * B );

X = x0 + K *;

Y = y0 + K * B;

· Intersection of known straight lines and straight lines (x, y)

If b1 = 0

Otherwise

AC code

# Include <iostream> # include <cstdio> using namespace STD; struct node {Double X, Y ;}; double A1, B1, C1, A2, B2, C2, K; node N1, N2, S, E, node1; int main () {int N; CIN> N; while (n --) {scanf ("% lf", & n1.x, & n1.y, & n2.x, & n2.y, & S. x, & S. y, & E. x, & E. y); {// locate the mirror line a1 = n2.y-n1.y; b1 = n1.x-n2.x; C1 = n1.y * n2.x-n1.x * n2.y; // obtain the symmetry point K =-2*(A1 * s. X + B1 * s. Y + C1)/(A1 * A1 + B1 * B1); node1.x = S. X + K * A1; node1.y = S. Y + K * B1; a2 = E. y-node1.y; b2 = node1.x-e. x; C2 = node1.y * E. x-node1.x * E. y; // calculate the intersection of two straight lines Double X, Y; If (b1 = 0) {x =-C1/A1; y = (-c2-a2 * X) /B2;} else {x = (C2 * b1-c1 * B2)/(A1 * b2-a2 * B1); y = (-A1 * x-c1)/B1 ;} printf ("%. 3lf %. 3lf \ n ", x, y) ;}} return 0 ;}

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