HDU 3934 凸包

题意:给定一些点，求最大三角形面积

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 1 /*

 2 凸包+最大三角形面积

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #include<algorithm>

 8 #include<iostream>

 9 #include<queue>

10 //#include<map>

11 #include<math.h>

12 using namespace std;

13 typedef long long ll;

14 //typedef __int64 int64;

15 const int maxn = 1000005;

16 const int inf = 0x7fffffff;

17 const double pi=acos(-1.0);

18 struct node{

19     int x,y;

20     bool operator <( const node & p ) const {

21         return y<p.y||(y==p.y&&x<p.x);

22     }

23 };

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

25 

26 int cross ( node sp,node ep,node op ){

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

28 }

29 

30 int dis2( node a,node b ){

31     return ( a.x-b.x )*( a.x-b.x )+( a.y-b.y )*( a.y-b.y );

32 }

33 

34 int garham( int n ){

35     int top=1;

36     sort( pnt,pnt+n );

37     if( n==0 ) return 0;

38     else res[ 0 ]=pnt[ 0 ];

39     if( n==1 ) return 1;

40     else res[ 1 ]=pnt[ 1 ];

41     if( n==2 ) return 2;

42     else res[ 2 ]=pnt[ 2 ];

43     

44     for( int i=2;i<n;i++ ){

45         while( top>0&&cross( res[ top-1 ],pnt[ i ],res[ top ] )>=0 )

46             top--;

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

48     }

49     

50     int len=top;

51     res[ ++top ]=pnt[ n-2 ];

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

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

54             top--;

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

56     }

57     

58     return top;

59 }

60 

61 int main(){

62     int n;

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

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

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

66         int cnt=garham( n );

67         double ans=0;

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

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

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

71                     ans=max( ans,( 0.5*cross( res[ j ],res[ k ],res[ i ] ) ) );

72                 }

73             }

74         }

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

76     }

77     return 0;

78 }