最近学习图形学,开始记录所学所长:
根据直线方程:F(x, y) = ax + by + c = 0 其中, a = y0 - y1, b = x1 - x0, c = x0y1 - x1y0。 将中点代入函数得: d = F(M) = F(xp + 1, yp + 0.5) = a(xp + 1) + b(yp + 0.5) + c 所以当d<0时, M在直线下方, 当d >= 0时, M在直线上方 采用增量法 d>=0时,取(xp+2, yp+0.5) 代入得:d1 = F(xp+2, yp+0.5) = a(xp+2)+b(yp+0.5)+c = d + a 故增量为a d<0时,取(xp+2, yp+1.5) = a(xp+2)+b(yp+1.5)+c = d + a + b 故增量为b d的初值:d0 = F(x0 + 1, y0 + 0.5) = a(x0 + 1) + b(y0 + 0.5) + c = ax0 + by0 + c + a + 0.5b = F(x0, y0) + a + 0.5b 因为F(x0, y0) = 0 所以d0 = a + 0.5b 只需要d的符号,所以用2d代替d 程序为: public void MidpointLine(Graphics g, int x0, int y0, int x1, int y1, int color) { int a, b, delta1, delta2, d, x, y; a = y0 - y1; b = x1 - x0; d = 2 * a + b; delta1 = 2 * a; delta2 = 2 * (a + b); x = x0; y = y0; g.setColor(color); g.drawLine(x, y, x, y); while(x < x1) { if( d < 0) { x ++; y ++; d += delta2; } else { x ++; d += delta1; } g.setColor(color); g.drawLine(x, y, x, y); } } 编译通过

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