POJ 3301 Texas Trip (三分求极限）

 

题意：给出许多点的坐标，用最小的正方形覆盖之。

题解：三分。注意精度····。公式神马的画个图推一下即解决

double mid1 = (left + right)/2;             

double mid2 = (mid1 + right)/2;         

相比 

double mid1 =left +(right - left)/3;
double mid2 = right - (right - left)/3;
要求更高的精度。

 

#include <cmath>
#include <iostream>
using namespace std;

#define MAX 33
#define eps 0.00000005
#define max(a,b) (a>b?a:b)
struct { double x, y; } p[MAX];

double cal1 ( int n, double d )
{
	double dis1 = 0.0, temp;
	for ( int i = 1; i < n; i++ )
	{
		for ( int j = i + 1; j <= n; j++ )
		{
			temp = fabs( (p[i].y-p[j].y) * sin(d) + (p[i].x-p[j].x) * cos(d));
			dis1 = max ( dis1, temp );
		}
	}
	return dis1;
}

double cal2 ( int n, double d )
{
	double dis2 = 0.0, temp;
	for ( int i = 1; i < n; i++ )
	{
		for ( int j = i + 1; j <= n; j++ )
		{
			temp = fabs( (p[i].y-p[j].y) * cos(d) - (p[i].x-p[j].x) * sin(d));
			dis2 = max ( dis2, temp );
		}
	}
	return dis2;
}
				

int main()
{
	int cs, n;
	scanf("%d",&cs); 
	while ( cs-- )
	{
		scanf("%d",&n);
		for ( int i = 1; i <= n; i++ )
			scanf("%lf%lf", &p[i].x, &p[i].y);
		
		double ans = 1000, low = 0, high = asin(1.0);
		while ( high - low > eps )
		{  
			double mid1 = low + ( high - low ) / 3;
			double mid2 = high - ( high - low ) / 3;
			double len1 = max ( cal1 ( n, mid1 ), cal2 ( n, mid1 ));
			double len2 = max ( cal1 ( n, mid2 ), cal2 ( n, mid2 ));
			
			if ( len1 < len2 )
			{
				ans = len1;
				high = mid2;
			}
			else
			{
				ans = len2;
				low = mid1;
			}
		}
		printf("%.2lf\n",ans*ans);
	}
	return 0;
}

 

下面的迭代法转自http://hi.baidu.com/liugoodness/blog/item/1aaebb16494b314320a4e9ad.html

#include <iostream>
#include <cstdio>
#include <cmath>

#define MAX 1000000000
#define MIN -1000000000
#define PI (asin(1.)*2)

using namespace std;

long x[50],y[50];

int main()
{
	long t, n, j, i;	
	double stepLen, from, num, left, right, up, down;
	double nx, ny, border, minBorder, minA, a, getSin, getCos;
	scanf("%ld",&t);

	while ( t-- )	
	{
		scanf("%ld",&n);
		for ( j = 0; j < n; j++ )
			scanf ("%ld%ld",&x[j], &y[j] );
		
		from = 0; num = 15; minBorder = MAX; stepLen = PI / 18; a=0;
		
		while ( num-- )
		{
			for ( i = 0; i < 10; i++ )
			{
				getSin = sin(a); getCos = cos(a);
				down = left = MAX;
				up = right = MIN;
			
				for ( j = 0; j < n; j++ )
				{
					nx = x[j] * getCos - y[j] * getSin;
					ny = y[j] * getCos + x[j] * getSin;
					
					if ( nx < left ) left = nx;
					if ( nx > right) right = nx;
					if ( ny < down ) down = ny;
					if ( ny > up ) up = ny;
				}
				
				border = right - left;
				border = ( up - down ) > border ? ( up - down ) : border;
				if ( border < minBorder ) minBorder = border, from = a;
				a += stepLen;
			}
			
			a = from - stepLen;
			stepLen /= 4.5;
		}
		
		printf("%.2lf\n",minBorder*minBorder);
		
	}
	return 0;
}