Poj 3301 Texas trip (trigger limit)

 

The coordinates of many points are given and covered with the smallest square.

Question: three points. Pay attention to precision ····. Draw a diagram of the formula Shenma to solve the problem.

Double mid1 = (left + right)/2;

Double mid2 = (mid1 + right)/2;

Compared

Double mid1 = left + (right-left)/3;
Double mid2 = right-(right-left)/3;
Higher accuracy is required.

 

#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;}

 

The following iteration method is transferred from 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;}