[UVa] 10310 - Dog and Gopher

題目網址: https://goo.gl/rww5i1

題意:
(from luckycat)


解法:
簡單水題，判斷距離。

TAG: ad hoc

注意:

程式碼:

	/**
	* Tittle: 10310 - Dog and Gopher
	* Author: Cheng-Shih, Wong
	* Date: 2015/06/16
	*/
	// include files
	#include <bits/stdc++.h>
	using namespace std;
	// definitions
	#define FOR(i,a,b) for( int i=(a),_n=(b); i<=_n; ++i )
	#define clr(x,v) memset( x, v, sizeof(x) )
	#define N 1005
	// declarations
	int n;
	double dx, dy, gx, gy;
	double hx[N], hy[N];
	// functions
	void input( void )
	{
	scanf( "%lf%lf%lf%lf", &gx, &gy, &dx, &dy );
	FOR( i, 1, n )
	scanf( "%lf%lf", &hx[i], &hy[i] );
	}
	double dis( double ax, double ay, double bx, double by )
	{
	double delx = (ax-bx);
	double dely = (ay-by);
	return sqrt( delx*delx+dely*dely );
	}
	void solve( void )
	{
	int hole = 0;
	FOR( i, 1, n ) {
	if( 2*dis(gx,gy,hx[i],hy[i]) <= dis(dx,dy,hx[i],hy[i]) ) {
	hole = i;
	break;
	}
	}
	if( hole )
	printf( "The gopher can escape through the hole at (%.3lf,%.3lf).\n", hx[hole], hy[hole] );
	else
	puts("The gopher cannot escape.");
	}
	// main function
	int main( void )
	{
	// input
	while( scanf( "%d", &n )==1 ) {
	input();
	solve();
	}
	return 0;
	}

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