Solution UVA: 10310 - Dog and Gopher

Problem	Verdict	Lang	Time	Best	Rank	Submit Time
| discuss10310 - Dog and Gopher	Accepted	C++11	0.000	0.000	452	1 mins ago

Suggest:

- Because the dog run better x2 than mouse 

=> mouse escape <=> distance(mouse -> hole) * 2 < distance(dog -> hole)

#include<bits/stdc++.h>

using namespace std;

double sqr(double x){

    return x*x;

}

double distance(double x1, double y1, double x2, double y2){

    return sqrt(sqr(x1-x2) + sqr(y1-y2));

}

int main(){

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

   

    int n;

    double x1, x2, y1, y2;

   

    while(cin >> n >> x1 >> y1 >> x2 >> y2){

        vector<pair<double, double>> a;

       

        while(n--){

            double x, y;    cin >> x >> y;

            a.push_back({x,y});

        }

       

        bool escape = false;

        for(auto p:a){

           

            double x = p.first;

            double y = p.second;

           

            double d1 = distance(x, y, x1, y1);

            double d2 = distance(x, y, x2, y2);

           

            if (d1*2 - d2 < 0.000001){

                escape = true;

                cout<<"The gopher can escape through the hole at ("<<fixed<<setprecision(3)<<x<<","<<y<<").\n";

                break;

            }

        }

       

        if (not escape)

            cout<<"The gopher cannot escape.\n";

    }

}