1069: [SCOI2007]最大土地面积 - BZOJ

Description

在某块平面土地上有N个点，你可以选择其中的任意四个点，将这片土地围起来，当然，你希望这四个点围成的多边形面积最大。
Input

第1行一个正整数N，接下来N行，每行2个数x,y，表示该点的横坐标和纵坐标。
Output

最大的多边形面积，答案精确到小数点后3位。
Sample Input
5
0 0
1 0
1 1
0 1
0.5 0.5
Sample Output
1.000

HINT

数据范围 n<=2000, |x|,|y|<=100000

先求凸包

然后枚举四边形的对角线，分别找到距离最远的两个点更新答案（这两个点都是单调的），总复杂度是O（n^2）的

哪里写挫了，超慢，总时间1600ms

 const
         maxn=;
 type
         point=record
           x,y:double;
         end;
 var
         a:array[..maxn]of point;
         n:longint;
         min:point;

 function cj(a,b,c:point):double;
 begin
         exit((a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y));
 end;

 procedure swap(var x,y:point);
 var
         t:point;
 begin
         t:=x;x:=y;y:=t;
 end;

 procedure sort(l,r:longint);
 var
         i,j:longint;
         y:point;
 begin
         i:=l;
         j:=r;
         y:=a[(l+r)>>];
         repeat
           while cj(y,a[i],min)< do
             inc(i);
           while cj(y,a[j],min)> do
             dec(j);
           if i<=j then
           begin
             swap(a[i],a[j]);
             inc(i);
             dec(j);
           end;
         until i>j;
         if i<r then sort(i,r);
         if j>l then sort(l,j);
 end;

 procedure init;
 var
         i:longint;
 begin
         read(n);
         min.x:=maxlongint;
         min.y:=maxlongint;
         for i:= to n do
           begin
             read(a[i].x,a[i].y);
             if (a[i].x<min.x) or ((a[i].x=min.x) and (a[i].y<min.y)) then min:=a[i];
           end;
         i:=n;
         while i> do
           begin
             if (a[i].x=min.x) and (a[i].y=min.y) then
               begin
                 a[i]:=a[n];
                 dec(n);
               end
             else dec(i);
           end;
         sort(,n);
         inc(n);
         a[n]:=min;
         a[]:=min;
 end;

 var
         q:array[..maxn]of longint;
         f:array[..maxn,..]of double;
         tot,lasta,lastb:longint;
         ans:double;

 function up(var x:double;y:double):boolean;
 begin
         if x<y then
         begin
           x:=y;
           exit(true);
         end;
         exit(false);
 end;

 function down(var x:double;y:double):boolean;
 begin
         if x>y then
         begin
           x:=y;
           exit(true);
         end;
         exit(false);
 end;

 procedure work;
 var
         i,j:longint;
 begin
         for i:= to n do
           begin
             while (tot>) and (cj(a[i],a[q[tot]],a[q[tot-]])>=) do
               dec(tot);
             inc(tot);
             q[tot]:=i;
           end;
         for i:= to tot- do
           begin
             f[i+,]:=-maxlongint;
             f[i+,]:=;
             for j:= to n do
               if up(f[i+,],cj(a[q[i+]],a[q[j]],a[q[i]])) then lasta:=j;
             lastb:=i;
             for j:=i+ to tot do
               begin
                 f[j,]:=cj(a[q[j]],a[q[lasta]],a[q[i]]);
                 f[j,]:=cj(a[q[j]],a[q[lastb]],a[q[i]]);
                 while up(f[j,],cj(a[q[j]],a[q[lasta mod tot+]],a[q[i]])) do
                   lasta:=lasta mod tot+;
                 while down(f[j,],cj(a[q[j]],a[q[lastb mod tot+]],a[q[i]])) do
                   lastb:=lastb mod tot+;
                 up(ans,f[j,]-f[j,]);
               end;
           end;
         writeln(ans/::);
 end;

 begin
         init;
         work;
 end.