Java Crossing.solveQuad方法代码示例

import org.apache.harmony.awt.gl.Crossing; //导入方法依赖的package包/类
/**
 * It checks up if there is intersection of the line (x1, y1) - (x2, y2) and
 * the quad curve (qx1, qy1) - (qx2, qy2) - (qx3, qy3). The parameters of the intersection
 * area saved to params array. Therefore the params size must be at learst 4.
 * @return The method returns the quantity of roots lied in the defined interval 
 */
public static int intersectLineAndQuad(double x1, double y1, double x2, double y2,
                                       double qx1, double qy1, double qx2, double qy2, 
                                       double qx3, double qy3, double[] params) {
    double[] eqn = new double[3];
    double[] t = new double[2];
    double[] s = new double[2];
    double dy = y2 - y1;
    double dx = x2 - x1;
    int quantity = 0;
    int count = 0;

    eqn[0] = dy * (qx1 - x1) - dx * (qy1 - y1);
    eqn[1] = 2 * dy * (qx2 - qx1) - 2 * dx * (qy2 - qy1);
    eqn[2] = dy * (qx1 - 2 * qx2 + qx3) - dx *(qy1 -2 * qy2 + qy3);
    
    if ((count = Crossing.solveQuad(eqn, t)) == 0) {
        return 0;
    }

    for (int i = 0; i < count; i++) {
        if (dx != 0) {
            s[i] = (quad(t[i], qx1, qx2, qx3) - x1) / dx;
        } else if (dy != 0) {
            s[i] = (quad(t[i], qy1, qy2, qy3) - y1) / dy;
        } else {
        	s[i] = 0.0;
        }
        if (t[i] >= 0 && t[i] <= 1 && s[i] >= 0 && s[i] <= 1) {
            params[2 * quantity] = t[i];
            params[2 * quantity + 1] = s[i];
            ++quantity;
        }
    }

    return quantity;
}
 

import org.apache.harmony.awt.gl.Crossing; //导入方法依赖的package包/类
/**
 * It checks up if there is intersection of the line (x1, y1) - (x2, y2) and
 * the quad curve (qx1, qy1) - (qx2, qy2) - (qx3, qy3). The parameters of
 * the intersection area saved to params array. Therefore the params size
 * must be at learst 4.
 * 
 * @return The method returns the quantity of roots lied in the defined
 *         interval
 */
public static int intersectLineAndQuad(double x1, double y1, double x2, double y2, double qx1, double qy1,
		double qx2, double qy2, double qx3, double qy3, double[] params) {
	double[] eqn = new double[3];
	double[] t = new double[2];
	double[] s = new double[2];
	double dy = y2 - y1;
	double dx = x2 - x1;
	int quantity = 0;
	int count = 0;

	eqn[0] = dy * (qx1 - x1) - dx * (qy1 - y1);
	eqn[1] = 2 * dy * (qx2 - qx1) - 2 * dx * (qy2 - qy1);
	eqn[2] = dy * (qx1 - 2 * qx2 + qx3) - dx * (qy1 - 2 * qy2 + qy3);

	if ((count = Crossing.solveQuad(eqn, t)) == 0) {
		return 0;
	}

	for (int i = 0; i < count; i++) {
		if (dx != 0) {
			s[i] = (quad(t[i], qx1, qx2, qx3) - x1) / dx;
		} else if (dy != 0) {
			s[i] = (quad(t[i], qy1, qy2, qy3) - y1) / dy;
		} else {
			s[i] = 0.0;
		}
		if (t[i] >= 0 && t[i] <= 1 && s[i] >= 0 && s[i] <= 1) {
			params[2 * quantity] = t[i];
			params[2 * quantity + 1] = s[i];
			++quantity;
		}
	}

	return quantity;
}
 

import org.apache.harmony.awt.gl.Crossing; //导入方法依赖的package包/类
public static int solveQuadratic(double eqn[], double res[]) {
    return Crossing.solveQuad(eqn, res);
}
 

import org.apache.harmony.awt.gl.Crossing; //导入方法依赖的package包/类
public static int solveQuadratic(double eqn[], double res[]) {
	return Crossing.solveQuad(eqn, res);
}
 

import org.apache.harmony.awt.gl.Crossing; //导入方法依赖的package包/类
/**
 * Finds the roots of the quadratic polynomial. This is accomplished by
 * finding the (real) values of x that solve the following equation:
 * eqn[2]*x*x + eqn[1]*x + eqn[0] = 0. The solutions are written into the
 * array res starting from the index 0 in the array. The return value tells
 * how many array elements have been written by this method call.
 * 
 * @param eqn
 *            an array containing the coefficients of the quadratic
 *            polynomial to solve.
 * @param res
 *            the array that this method writes the results into.
 * @return the number of roots of the quadratic polynomial.
 * @throws ArrayIndexOutOfBoundsException
 *             if {@code eqn.length} < 3 or if {@code res.length} is less
 *             than the number of roots.
 * @throws NullPointerException
 *             if either array is null.
 */
public static int solveQuadratic(double eqn[], double res[]) {
    return Crossing.solveQuad(eqn, res);
}