ENG  RUSTimus Online Judge
Online Judge
Problems
Authors
Online contests
About Online Judge
Frequently asked questions
Site news
Webboard
Links
Problem set
Submit solution
Judge status
Guide
Register
Update your info
Authors ranklist
Current contest
Scheduled contests
Past contests
Rules
back to board

Discussion of Problem 1348. Goat in the Garden 2

AC with C++
Posted by Varun Sharma 21 Apr 2009 06:18
#include <iostream>
#include <cmath>

using namespace std;

int main(){
    double x1 = 0, y1 = 0, x2 = 0, y2 = 0;
    double px = 0, py = 0;
    double length;

    double A = 0;
    double B = -1;
    double C = 0;
    double slope = 0;

    cin>>x1>>y1>>x2>>y2>>px>>py>>length;

    // get the equation of pineapple line in the form Ax + By + c = 0

    if((x2 - x1) != 0){
        slope = (y2 - y1) / (x2 - x1);
        A = slope;
        C = y2 - slope * x2;
    }

    if((x2 - x1) == 0){
        A = 1;
        B = 0;
        C = -x1;
    }

    // now since we have Ax + By + C = 0
    // we can find the shortest distance between the point and the line

    double short_dist = fabs(A*px + B*py + C) / sqrt(pow(A, 2) + pow(B, 2));
    double dist_x1y1 = fabs(sqrt(pow(px-x1, 2) + pow(py-y1, 2)));
    double dist_x2y2 = fabs(sqrt(pow(px-x2, 2) + pow(py-y2, 2)));

    // once we have calculated the shorted distance, we need to
    // find out the point lying on that pineapple line which is
    // closest to the peg

    double slope_new = 0;
    double B_new = 0;
    double C_new = 0;
    double A_new = 0;
    bool status = true;

    if((x2 - x1) == 0){
        slope_new = 0; // the other line has slope of inifinity, so this is parallel
        B_new = 1;
        A_new = 0;
        C_new = -py;
        status = false;
    }

    if((y2 - y1) == 0){
        B_new = 0;
        A_new = 1;
        C_new = -px;
        status = false;
    }

    if(status){
        slope_new = -1.0/slope;
        C_new = py - slope_new*px;
        A_new = slope_new;
        B_new = -1;
    }

    // calculating the point of intersection

    double inter_X = (-C*B_new - B*(-C_new)) / (A*B_new - B*A_new);
    double inter_Y = (A*(-C_new) - (-C*A_new))/(A*B_new - B*A_new);

    // now check whether this point lies between the given points or not ?
    // to do that we need to check whether both of the above calculated
    // coordinates lie between the range of x1, y1 and x2, y2

    double min_x_coor = min(x1, x2);
    double min_y_coor = min(y1, y2);
    double max_x_coor = max(x1, x2);
    double max_y_coor = max(y1, y2);

    if(inter_X >= min_x_coor && inter_X <= max_x_coor && inter_Y >= min_y_coor && inter_Y <= max_y_coor){
        // this means, the point lies on the line
        double one_pineapple_distance = short_dist - length;
        if(one_pineapple_distance <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", one_pineapple_distance);

        // now to eat all the pineapples
        double larger_distance = max(dist_x1y1, dist_x2y2);
        if(larger_distance - length <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", larger_distance - length);
    }

    // or else the perpendicuar point does not lies on the line
    else{
        // first shorter one
        double shorter_distance = min(dist_x1y1, dist_x2y2);
        double larger_distance = max(dist_x1y1, dist_x2y2);

        double pineapple_1 = shorter_distance - length;
        if(pineapple_1 <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", pineapple_1);

        double pineapple_all = larger_distance - length;
        if(pineapple_all <= 0){
            cout<<"0.00"<<endl;
        }
        else
            printf("%.2f\n", pineapple_all);
    }
    return 0;
}
Re: AC with C++
Posted by kwetril 25 Dec 2010 01:47
вай, как много кода)
с помощью численных методов решение в 30 строк AC)
Re: AC with C++
Posted by mikroz 16 Oct 2013 22:41
С помощью точной формулы решение в 25 строк.
Re: AC with C++
Posted by vfursov@sfedu.ru 24 May 2024 16:55