HDU2202 凸包

凸包算法+枚举

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 1  /*

 2  凸包

 3  顺时针！！！！

 4  */

 5  #include<stdio.h>

 6  #include<string.h>

 7  #include<stdlib.h>

 8  #include<algorithm>

 9  #include<iostream>

10  #include<queue>

11  #include<stack>

12  #include<math.h>

13  #include<map>

14  using namespace std;

15  const int maxn = 50005;

16  struct node{

17      int x,y;

18  };

19  bool operator < ( const node &l,const node &r ){

20          return l.y<r.y||( l.y==r.y&&l.x<r.x );

21  }

22  node pnt[ maxn ],res[ maxn ];

23 int mult( node sp,node ep,node op ){

24      return (sp.x - op.x) * (ep.y - op.y)-(ep.x - op.x) * (sp.y - op.y);

25  }

26  /*

27  ep

28  |

29  |

30  op----sp

31  ( from sp to ep )

32  */

33  int graham( int n ){

34      int i, len, k = 0;

35      int top = 1;

36      sort(pnt, pnt + n);

37      if (n == 0) return 0; res[0] = pnt[0];

38      if (n == 1) return 1; res[1] = pnt[1];

39      if (n == 2) return 2; res[2] = pnt[2];

40      for (i = 2; i < n; i++) {

41          while (top && mult(res[ top ], pnt[ i ], res[top-1])>=0 )//( cross ： from top to i )

42              top--;

43          res[++top] = pnt[i];

44      }

45      len = top; res[++top] = pnt[n - 2];

46      for (i = n - 3; i >= 0; i--) {

47          while (top!=len && mult(res[ top ], pnt[ i ], res[top-1])>=0 ) 

48          top--;

49          res[++top] = pnt[i];

50      }

51      return top; // 返回凸包中点的个数

52  }

53  int main(){

54      int n;

55      while( scanf("%d",&n)!=EOF ){

56          for( int i=0;i<n;i++ )

57              scanf("%d%d",&pnt[i].x,&pnt[i].y);

58          int cnt=graham( n );

59          double ans=0;

60          for( int i=0;i<cnt;i++ ){

61              for( int j=i+1;j<cnt;j++ ){

62                  for( int k=j+1;k<cnt;k++ ){

63                      double tmp=mult( res[i],res[k],res[j] );

64                      ans=max( ans,tmp );

65                  }

66              }

67          }

68          ans*=0.5;

69          printf("%.2lf\n",ans);

70      }

71      return 0;

72  }