Summer holiday, 1005, 2011 Multi-University Training Contest 10

Summer holiday, 1005, 2011 Multi-University Training Contest 10

Summer holiday

TimeLimit: 1 Second   MemoryLimit: 32 Megabyte

Totalsubmit: 434   Accepted: 108  

Description

Summer holiday was coming! Xiaomao went back to his hometown where he yearn day and night, his hometown has picturesque scenery. There is a big forest beside his village. There are n trees in the forest.
Now they want to across the forest with a rope (the rope won't cross). Try to find 3 trees in this tree on the rope which can make the area of the surrounded largest. Work out the area of it.

Input

The input will consist of several test cases. The first line contains a positive integer N（3<=N<=10^6）, the number of trees, followed N lines, each gives the (xi, yi ) coordinates.

Output

Print the largest area, one number a line with two decimal places.

Sample Input

4
0 0
1 1
0 1
1 0

Sample Output

0.50

Source

[p][/p]

二维凸包

不做 ACM 三个月了，心血来潮参加了练习赛，悲剧的没有准备模板，这个模板是临时从网上搜来的，非原创。

  1  #include < iostream >
  2  #include < cstdio >
  3  #include < cmath >
  4  #include < cstdlib >
  5  #include < algorithm >
  6 
  7  using   namespace  std;
  8 
  9  struct  P{
 10           double  x,y;
 11  };
 12 
 13  #define   EPS  0.00001
 14  #define   ZERO(x)   ( (x<EPS) && ((-(x))<EPS) )
 15 
 16  const   int  L  =   2000009 ;
 17  P p[ L ], stack[ L ];
 18  int  n, top;
 19 
 20  inline  double  Mul(P p1,P p2,P p3) 
 21  {    
 22           return  (p2.x - p1.x) * (p3.y - p1.y) - (p2.y - p1.y) * (p3.x - p1.x); 
 23  }
 24 
 25  inline  double  dis(P a,P b)
 26  {
 27           return  sqrt((a.x - b.x) * (a.x - b.x)  +  (a.y - b.y) * (a.y - b.y));
 28  }
 29 
 30  int  cmp( const   void   * a, const   void   * b)
 31  {
 32          P  *  c  =  (P  * )a;
 33          P  *  d  =  (P  * )b;
 34           double  k  =  Mul(p[ 0 ], * c, * d);
 35           if (k < 0   ||  ( ! k  &&  dis( * c,p[ 0 ])  >  dis( * d,p[ 0 ]) ) )
 36                   return   1 ;
 37           return   - 1 ;
 38  }
 39 
 40  inline  void  tubao( int  n, int   & top)
 41  {
 42           int  i;
 43          top  =   2 ;
 44          stack[ 0 ]  =  p[ 0 ];
 45          stack[ 1 ]  =  p[ 1 ];
 46          stack[ 2 ]  =  p[ 2 ];
 47           for (i = 3 ;i <= n;i ++ )
 48          {
 49                   while (Mul(stack[top - 1 ],stack[top],p[i]) <= 0   &&  top >= 2 )
 50                          top  -- ;
 51                  top  ++ ;
 52                  stack[top]  =  p[i];
 53          }
 54  }
 55 
 56  inline  double  displ( P p, P l0, P l1 ) {
 57           double  t  =  ( (p.x - l0.x) * (l1.x - l0.x)  +  (p.y - l0.y) * (l1.y - l0.y) )  /  ( dis(l0,p)  *  dis(l0,l1) );
 58           return  dis(p,l0)  *  sqrt(  1   -  t  *  t );
 59  }
 60 
 61  inline  double  area( P a, P b, P c ) {
 62           return  dis(a,b)  *  displ(c,a,b)  /   2 ;
 63  }
 64 
 65  double  solve() {
 66           int  i, j, k;
 67           double  ans  =   0 , anstmp;
 68           for  ( i  =   0 ; i  <  top;  ++ i ) {
 69               for  ( j  =  i  +   1 ; j  <  top;  ++ j ) {
 70                   for  ( k  =  j  +   1 ; k  <  top;  ++ k ) {
 71                      anstmp  =  area( stack[ i ], stack[ j ], stack[ k ] );
 72                       if  ( anstmp  >  ans ) {
 73                          ans  =  anstmp;
 74                      }
 75                  }
 76              }
 77          }
 78           return  ans;
 79  }
 80 
 81  int  main()
 82  {
 83           int  i,tar;
 84           double  x,y;
 85          P temp;
 86           while ( scanf( " %d " , & n)  ==   1 ) {
 87                  tar  =   0 ;
 88                  x  =  y  =   0x7FFFFFFF ;
 89                   for (i = 0 ;i < n;i ++ )
 90                  {
 91                          scanf( " %lf %lf " , & p[i].x, & p[i].y);
 92                           if (p[i].x < x  ||  p[i].x == x  &&  p[i].y < y)
 93                          {
 94                                  x  =  p[i].x;
 95                                  y  =  p[i].y;
 96                                  tar  =  i;
 97                          }
 98                  }
 99                  temp  =  p[tar];
100                  p[tar]  =  p[ 0 ];
101                  p[ 0 ]  =  temp;
102                  qsort(p + 1 ,n - 1 , sizeof (p[ 0 ]),cmp);
103                  p[n]  =  p[ 0 ];
104                  tubao(n,top);
105                  printf(  " %0.2lf\n " , solve() );
106          }
107           return   0 ;
108  }
109