poj1379 run away

题目大意：在矩形中找到一个点，使其跟最近的点距离最大

算法：模拟退火

 

模拟退火博大精深。。。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#define max( x , y ) ( ( x ) > ( y ) ? ( x ) : ( y ) )
#define min( x , y ) ( ( x ) < ( y ) ? ( x ) : ( y ) )
#define FOR( TMP , ST , ED ) for ( int TMP = ( ST ) ; TMP < ( ED ) ; TMP ++ )
#define RORD( TMP , ST , ED ) for ( int TMP = ( ST ) ; TMP > ( ED ) ; TMP ++ )
const int max_int = ( 2147483467 ) , oo = ( 1e9 ) ;
using namespace std ;

const int maxn = ( 1e3 + 10 ) , Num = ( 30 ) ;
int N , Kuan , Chang ;
double X[ Num ] , Y[ Num ] , Dis[ Num ] , data[ maxn ][ 2 ] ;

#define sqr( x ) ( ( x ) * ( x ) )
inline double Get( double ST_x , double ST_y )
{
	double ans = ( double ) oo ;
	FOR ( i , 0 , N )
		ans = min( ans , 
			sqrt( sqr( ST_x - data[ i ][ 0 ] ) + sqr( ST_y - data[ i ][ 1 ] ) ) ) ;
	
	return ans ;
}

int main()
{
	freopen( "poj1379.in" , "r" , stdin ) , freopen( "poj1379.out" , "w" , stdout ) ;
	int Test ;
	srand( time( 0 ) ) ;
	
	for ( scanf( "%d" , &Test ) ; Test -- ; )
	{
		cout << "The safest point is " ;
		scanf( "%d%d%d" , &Kuan , &Chang , &N ) ;
		FOR ( i , 0 , N )
			scanf( "%lf%lf" , &data[ i ][ 0 ] , &data[ i ][ 1 ] ) ;
		
		
		FOR ( i , 0 , Num )
		{
			X[ i ] = rand() % Kuan + 0.01 , Y[ i ] = rand() % Chang + 0.01 ;
			Dis[ i ] = Get( X[ i ] , Y[ i ] ) ;
		}
		
		double T = ( double ) max( Kuan , Chang ) / sqrt( ( double ) Num ) , Dec = 0.80 ;
		double eps = ( 1e-7 ) , Pi = asin( 1.0 ) * 2.0 ;
		for ( ; T >= eps ; T *= Dec )
			FOR ( i , 0 , Num )
				FOR ( j , 0 , Num )
				{
					double xx , yy , New , Yzx = ( ( double ) rand() / 32768.0 ) * 2.0 * Pi ;
					xx = X[ i ] + cos( Yzx ) * T , yy = Y[ i ] + sin( Yzx ) * T ;
					New = Get( xx , yy ) ;
					
						if ( New > Dis[ i ] && 
						xx >= 0 && xx <= Kuan && yy >= 0 && yy <= Chang )
							Dis[ i ] = New , X[ i ] = xx , Y[ i ] = yy ;
				}
		
		double Ans = ( double ) - oo , xx , yy ;
		FOR ( i , 0 , Num )
			if ( Ans < Dis[ i ] )
				Ans = Dis[ i ] , xx = X[ i ] , yy = Y[ i ] ;
		
		printf( "(%.1lf, %.1lf).\n" , xx , yy ) ;
	}
	
	return 0 ;
}