POJ3301 Texas Trip

这题的大意是在整数坐标上，给定一些点，问说覆盖这些点的最小的正方形的面积，好像黑书上有道类似的，把正方形改成矩形，对于那道题的解法好像是，考虑到最后矩形至少会有一条边上有两个以上的点，所以枚举两个点，可能生成这个矩形，这样就把可能旋转的角度给离散化了，并且复杂度只有O(N^2)的，或者可能都有旋转卡壳的方法来解这个。但对于这道题，不知道这样离散可不可以，但想法都是把枚举数量减少下来，这里用到逼近迭代，就是先m分枚举范围，从中找到最小的，然后再在那个区间m分，这样迭代下去。

感谢：

http://hi.baidu.com/liugoodness/blog/item/1aaebb16494b314320a4e9ad.html
http://hi.baidu.com/czyuan_acm/blog/item/8cc45b1f30cefefde1fe0b7e.html

代码

   
     

    
      
    
      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      string
    
      >
    
      
#include 
    
      <
    
      vector
    
      >
    
      
#include 
    
      <
    
      map
    
      >
    
      
#include 
    
      <
    
      set
    
      >
    
      
#include 
    
      <
    
      queue
    
      >
    
      
#include 
    
      <
    
      math.h
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      const
    
       
    
      int
    
       MAX 
    
      =
    
       
    
      305
    
      ;

    
      const
    
       
    
      double
    
       PI 
    
      =
    
       asin(
    
      1.0
    
      ) 
    
      *
    
       
    
      2
    
      ;

    
      const
    
       
    
      double
    
       EPS 
    
      =
    
       1e
    
      -
    
      6
    
      ;


    
      int
    
       n;

    
      double
    
       x[MAX];

    
      double
    
       y[MAX];

    
      double
    
       xx[MAX];

    
      double
    
       yy[MAX];


    
      double
    
       check(
    
      double
    
       a)
{
 
    
      double
    
       minx 
    
      =
    
       
    
      1000000
    
      ;
 
    
      double
    
       maxx 
    
      =
    
       
    
      -
    
      minx;
 
    
      double
    
       miny 
    
      =
    
       minx;
 
    
      double
    
       maxy 
    
      =
    
       maxx;
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      )
 {
 xx[i] 
    
      =
    
       cos(a) 
    
      *
    
       x[i] 
    
      +
    
       sin(a) 
    
      *
    
       y[i];
 yy[i] 
    
      =
    
       
    
      -
    
      sin(a) 
    
      *
    
       x[i] 
    
      +
    
       cos(a) 
    
      *
    
       y[i];
 minx 
    
      =
    
       min(minx, xx[i]);
 maxx 
    
      =
    
       max(maxx, xx[i]);
 miny 
    
      =
    
       min(miny, yy[i]);
 maxy 
    
      =
    
       max(maxy, yy[i]);
 }
 
    
      double
    
       e 
    
      =
    
       (maxx 
    
      -
    
       minx) 
    
      >
    
       (maxy 
    
      -
    
       miny) 
    
      ?
    
       (maxx 
    
      -
    
       minx) : (maxy 
    
      -
    
       miny);
 
    
      return
    
       e 
    
      *
    
       e;
}


    
      double
    
       go()
{
 
    
      int
    
       m 
    
      =
    
       
    
      800
    
      ;
 
    
      int
    
       times 
    
      =
    
       
    
      60
    
      ;
 
    
      double
    
       begin 
    
      =
    
       
    
      0
    
      ;
 
    
      double
    
       end 
    
      =
    
       PI 
    
      /
    
       
    
      3
    
      ;
 
    
      double
    
       res 
    
      =
    
       
    
      -
    
      1
    
      ;
 
    
      double
    
       from;

 
    
      double
    
       a; 
 
    
      double
    
       da;

 
    
      while
    
      (times
    
      --
    
      )
 {
 a 
    
      =
    
       begin;
 da 
    
      =
    
       (end 
    
      -
    
       begin) 
    
      /
    
       m;

 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       m; i
    
      ++
    
      )
 {
 
    
      double
    
       t 
    
      =
    
       check(a 
    
      +
    
       da 
    
      *
    
       i);
 
    
      if
    
      (res 
    
      ==
    
       
    
      -
    
      1
    
       
    
      ||
    
       t 
    
      <
    
       res)
 {
 res 
    
      =
    
       t;
 from 
    
      =
    
       a 
    
      +
    
       da 
    
      *
    
       i;
 }
 }

 begin 
    
      =
    
       from 
    
      -
    
       da;
 end 
    
      =
    
       from 
    
      +
    
       
    
      2
    
       
    
      *
    
       da;
 }

 
    
      return
    
       res;
}


    
      int
    
       main()
{
 
    
      int
    
       T;
 scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      T);
 
    
      while
    
      (T
    
      --
    
      )
 {
 scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n);
 
    
      for
    
      (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       n; i
    
      ++
    
      ) scanf(
    
      "
    
      %lf%lf
    
      "
    
      , 
    
      &
    
      x[i], 
    
      &
    
      y[i]);
 printf(
    
      "
    
      %.2lf\n
    
      "
    
      , go());
 }
}