Sicily 2376. Cutting The Cake

2376. Cutting The Cake

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB , Special Judge

Description

 

For each birthday of twins James and John, their mother bakes a special triangular birthday cake with a different shape each year. The sides of the cake are frosted with chocolate icing and the top is frosted with special pink icing. James and John both like both types of icing and insist that they get their equal share of each.

Write a program to help James and John cut the cake so they each get an equal share. A triangle equalizer is a line through a triangle that simultaneously divides the perimeter and area of the triangle into two equal pieces. If the triangle is isosceles, the bisector of the angle between two equal sides is such a line (A and B below). In general, a triangle may have1, 2 or 3 such lines (2 requires special conditions not likely to be found in a random triangle).

Write a program to compute the equation of one such dividing line (it does not matter which one) for a non-degenerate input triangle. You should use double-precision calculations for intermediate values.

Input

The first line of input contains a single integer P, (1P1000), which is the number of data sets that follow. Each data set consists of a single line containing a single decimal integer and 6 floating point numbers. The integer is the data set number (starting at 1). The floating-point numbers are the coordinates of the vertices of the input triangle: x₀, y₀, x₁, y₁, x₂, y₂.

Output

 

For each data set, there is a single line of output. The line contains a decimal integer giving the data set number followed by a single space, followed by 3 space-separated floating-point numbers to 5 decimal places. The 3 floating-point values are the coefficients (A, B and C) of the equation of the equalizer line:

A * x + B * y = C

where A * A + B * B = 1.0 and A 0.

Sample Input

4 1 0 0 4 10 8 0 2 0 0 10 8 8 0 3 0 0 -8 4.5 5 04 0 0 -5 4.6 5 0

Sample Output

1 1.00000 0.00000 4.00000 2 0.99347 -0.11408 5.98771 3 0.45018 0.89294 0.81417 4 0.66369 0.74801 1.01564

Problem Source

每周一赛第一场

// Problem#: 2376// Submission#: 3593630// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University/* * I - Cutting The Cake * ACM International Collegiate Programming Contest * Greater New York Region * October 24, 2010 */#include <stdio.h>#include <math.h>#define EPS (.00001)/* * (below <XY> = vector from point X to point Y and |V| = magnitude of vector V) * we have a triangle with vertices A, B, C with the length of * the side opposite A, B or C being a, b or c, respectively * a line intersects the triangle on side AB at a fraction x (0<= x<= 1) * of the distance from A to B and intersects the side AC at a fraction * y of the distance from A to C *               A *              / \ *          x*c/    \  y*b *            /       \ *         ----------------------- *          /            \ *         B______________C * * the area of the triangle: Area = 1/2 |<AB> X <AC>| (cross product) * the area of the triangle cut off by the line is * 1/2 | (x *<AB>) X (y * <AC>)| = 1/2*x*y * Area (of entire triangle) * perimeter of entire triangle = a+b+c * part of perimeter cut off by the line is x*c + y*b * * If the line is an equalizer for the triangle: * cutoff area = 1/2 triangel area => x*y = 1/2 * cutoff perimeter = 1/2 trinagle perimeter => * x*c + y*b = 1/2(a+b+c) * Solving: *  y = ((1/2)(a+b+c) - x*c)/b *  x * (((1/2)(a+b+c) - x*c)/b = 1/2 * clear denominators: *  x*(a+b+c) - 2*c*x*x = b OR *  2*c*x^2 - (a+b+c)*x + b = 0; *  x = [(a+b+c) +- sqrt((a+b+c)^2 - 8*b*c)]/4*c *  y = [(a+b+c) -+ sqrt((a+b+c)^2 - 8*b*c)]/4*b * * we need to try this for each choice of cut off vertex A, B or C * to find a case where ((a+b+c)^2 - 8*b*c) >= 0 AND * x and y are between 0 and 1 (including 1) * * when found we have two points on the equalizer line * (x0,y0) = A + x * <AB> * (x1,y1) + A + y * <AC> * the equation of the line is * (y1 - y0)*X + (x1- x0)*Y = (y1 - y0)*x0 + (x1- x0)*y0 *//* * check solns of *  x = [(a+b+c) +- sqrt((a+b+c)^2 - 8*b*c)]/4*c *  y = [(a+b+c) -+ sqrt((a+b+c)^2 - 8*b*c)]/4*b * return number of real solns with 0 < x,y <= 1 * pSoln should have 4 slots * pSoln[0] = first soln x   pSoln[1] = first soln y  * pSoln[2] = second soln x   pSoln[3] = second soln y  */int CheckEqSolns(double a, double b, double c, double *pSoln){    double perim, discr, x, y;    int ret = 0;    perim = a+b+c;    discr = perim*perim - 8*b*c;    if(discr < 0) {        return 0;    }    discr = sqrt(discr);    x = (perim + discr)/(4.0*c);    y = (perim - discr)/(4.0*b);    if((x > 0.0) && (x <= (1.0 + EPS)) && (y > 0.0) && (y <= (1.0 + EPS))) {        ret++;        pSoln[0] = x;        pSoln[1] = y;    }    x = (perim - discr)/(4.0*c);    y = (perim + discr)/(4.0*b);    if((x > 0.0) && (x <= (1.0 + EPS)) && (y > 0.0) && (y <= (1.0 + EPS))) {        pSoln[2*ret] = x;        pSoln[2*ret+1] = y;        ret++;    }    return ret;}int FindEqualizer(double *pVerts, double *pResult){    double vAB[2], vAC[2], vBC[2], soln[4];    double a, b, c, x0, y0, x1, y1, denom;    vAB[0] = pVerts[2] - pVerts[0];    vAB[1] = pVerts[3] - pVerts[1];    vAC[0] = pVerts[4] - pVerts[0];    vAC[1] = pVerts[5] - pVerts[1];    vBC[0] = pVerts[4] - pVerts[2];    vBC[1] = pVerts[5] - pVerts[3];    a = sqrt(vBC[0]*vBC[0] + vBC[1]*vBC[1]);    b = sqrt(vAC[0]*vAC[0] + vAC[1]*vAC[1]);    c = sqrt(vAB[0]*vAB[0] + vAB[1]*vAB[1]);    if(fabs(a-b) < EPS) {   // isoceles vertes at c        x0 = pVerts[4];        y0 = pVerts[5];        x1 = 0.5*(pVerts[0] + pVerts[2]);        y1 = 0.5*(pVerts[1] + pVerts[3]);    }    else if(fabs(a-c) < EPS) {  // isoceles vertes at b        x0 = pVerts[2];        y0 = pVerts[3];        x1 = 0.5*(pVerts[0] + pVerts[4]);        y1 = 0.5*(pVerts[1] + pVerts[5]);    }    else if(fabs(b-c) < EPS) {  // isoceles vertes at b        x0 = pVerts[0];        y0 = pVerts[1];        x1 = 0.5*(pVerts[2] + pVerts[4]);        y1 = 0.5*(pVerts[3] + pVerts[5]);    }    else if(CheckEqSolns(a, b, c, &(soln[0])) > 0) { // soln crossin AB and AC        x0 = pVerts[0] + vAB[0]*soln[0];        y0 = pVerts[1] + vAB[1]*soln[0];        x1 = pVerts[0] + vAC[0]*soln[1];        y1 = pVerts[1] + vAC[1]*soln[1];    }    else if(CheckEqSolns(b, c, a, &(soln[0])) > 0) { // soln crossing BC and BA        x0 = pVerts[2] + vBC[0]*soln[0];        y0 = pVerts[3] + vBC[1]*soln[0];        x1 = pVerts[2] - vAB[0]*soln[1];        y1 = pVerts[3] - vAB[1]*soln[1];    }    else if(CheckEqSolns(c, a, b, &(soln[0])) > 0) { // soln crossing CA and CB        x0 = pVerts[4] - vAC[0]*soln[0];        y0 = pVerts[5] - vAC[1]*soln[0];        x1 = pVerts[4] - vBC[0]*soln[1];        y1 = pVerts[5] - vBC[1]*soln[1];    }    else {        return -1;    }    a = y1 - y0;    b = -(x1-x0);    if(a < 0.0) {        a = -a;        b = -b;    }    denom = sqrt(a*a + b*b);    pResult[0] = a/denom;    pResult[1] = b/denom;    pResult[2] = (a*x0 + b*y0)/denom;    return 0;}char inbuf[256];int main(){    int nprob, curprob, index;    double invals[6], result[3];    if(fgets(&(inbuf[0]), 255, stdin) == NULL)    {        fprintf(stderr, "Read failed on problem count\n");        return -1;    }    if(sscanf(&(inbuf[0]), "%d", &nprob) != 1)    {        fprintf(stderr, "Scan failed on problem count\n");        return -2;    }    for(curprob = 1; curprob <= nprob ; curprob++)    {        if(fgets(&(inbuf[0]), 255, stdin) == NULL)        {            fprintf(stderr, "Read failed on problem %d data\n", curprob);            return -3;        }        if(sscanf(&(inbuf[0]), "%d %lf %lf %lf %lf %lf %lf", &index,            &(invals[0]), &(invals[1]), &(invals[2]),            &(invals[3]), &(invals[4]), &(invals[5])) != 7)        {            fprintf(stderr, "Scan failed on problem %d data\n", curprob);            return -4;        }        if(index != curprob)        {            fprintf(stderr, "problem index %d not = expected problem %d\n",                index, curprob);            return -7;        }        if(FindEqualizer(&(invals[0]), &(result[0])) == 0) {            printf("%d %.5f %.5f %.5f\n", curprob,                result[0], result[1], result[2]);        } else {            printf("no soln found\n");        }    }    return 0;}