Poj 3301 Texas trip

Question:

Points are given in the two-dimensional coordinate system, and the area of the smallest square can be covered (the side of a square is not necessarily a parallel coordinate axis)

Solution:

For a point, if the coordinate axis rotates degrees A (in radians), then x' = x * Cos (a)-y * sin (); y' = y * Cos (A) + x * sin ();

For the three-point angle, the square area is a single-peak function in [0, Pi ]. Has a minimum value.

The following code is used:

 

#include <set>#include <map>#include <queue>#include <math.h>#include <vector>#include <string>#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#define eps 1e-8#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr<<1#define rc m + 1,r,tr<<1|1#define iabs(x)  ((x) > 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (SIZE))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )#define min( x, y )  ( ((x) < (y)) ? (x) : (y) )using namespace std;struct node{    double x,y;}point[305];int n;double does(double a){    double maxx=-10000000,maxy=-100000000,minx=100000000,miny=100000000,tx,ty;    for(int i=0;i<n;i++)    {        tx=point[i].x*cos(a)-point[i].y*sin(a);        ty=point[i].y*cos(a)+point[i].x*sin(a);        maxx=max(maxx,tx);        maxy=max(maxy,ty);        minx=min(minx,tx);        miny=min(miny,ty);    }    return max(maxx-minx,maxy-miny);}int main(){    int T;    double l,r,lmid,rmid,ans1,ans2;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&point[i].x,&point[i].y);        }        l=0.0;        r=pi;        while(r-l>eps)        {            lmid=(l+r)/2;            rmid=(lmid+r)/2;            ans1=does(lmid);            ans2=does(rmid);            if(ans1<=ans2)r=rmid;            else l=lmid;        }        printf("%.2lf\n",ans1*ans1);    }    return 0;}