poj2187

求最远点对，这是一道经典的旋转卡壳的题目
话说由于是前年写的，之后就没怎么研究过计算几何了……
感觉都不大记得清了，来稍微回忆一下……
首先最远点对一定出现在凸包上显然，然后穷举肯定不行，这时候就需要旋转卡壳了
http://www.cnblogs.com/Booble/archive/2011/04/03/2004865.html 这里面介绍的非常详细

 var q,x,y:array[..] of longint;
     p,ans,i,j,h,r,k,t,n:longint;

 function cross(i,j,k,r:longint):longint;
   begin
     exit((x[i]-x[j])*(y[k]-y[r])-(x[k]-x[r])*(y[i]-y[j]));
   end;

 function dis(i,j:longint):longint;
   begin
     exit(sqr(x[i]-x[j])+sqr(y[i]-y[j]));
   end;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure sort(l,r: longint);
   var i,j,p,q: longint;
   begin
     i:=l;
     j:=r;
     p:=x[(l+r) shr ];
     q:=y[(l+r) shr ];
     repeat
       while (x[i]<p) or (x[i]=p) and (y[i]<q) do inc(i);
       while (p<x[j]) or (p=x[j]) and (q<y[j]) do dec(j);
       if not(i>j) then
       begin
         swap(x[i],x[j]);
         swap(y[i],y[j]);
         inc(i);
         j:=j-;
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;
 begin
   readln(n);
   for i:= to n do
     readln(x[i],y[i]);
   sort(,n);
   q[]:=;
   t:=;
   for i:= to n do
   begin
     while (t>) and (cross(q[t],q[t-],i,q[t-])>=) do dec(t);
     inc(t);
     q[t]:=i;
   end;
   k:=t;
   for i:=n- downto  do
   begin
     while (t>k) and (cross(q[t],q[t-],i,q[t-])>=) do dec(t);
     inc(t);
     q[t]:=i;
   end;
   h:=;
   r:=k;
   ans:=dis(h,r);
   while (h<=k) and (r<=t) do
   begin
     p:=cross(q[h],q[h+],q[r+],q[r]);  //两条直线谁先贴住凸包，这个可以用叉积解决，具体画个图就明白了
     if p<= then inc(h) else inc(r);
     ans:=max(ans,dis(q[h],q[r]));
   end;
   writeln(ans);
 end.