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[0..50010] 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 1];
    q:=y[(l+r) shr 1];
    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-1;
      end;
    until i>j;
    if l<j then sort(l,j);
    if i<r then sort(i,r);
  end;
begin
  readln(n);
  for i:=1 to n do
    readln(x[i],y[i]);
  sort(1,n);
  q[1]:=1;
  t:=1;
  for i:=2 to n do
  begin
    while (t>1) and (cross(q[t],q[t-1],i,q[t-1])>=0) do dec(t);
    inc(t);
    q[t]:=i;
  end;
  k:=t;
  for i:=n-1 downto 1 do
  begin
    while (t>k) and (cross(q[t],q[t-1],i,q[t-1])>=0) do dec(t);
    inc(t);
    q[t]:=i;
  end;
  h:=1;
  r:=k;
  ans:=dis(h,r);
  while (h<=k) and (r<=t) do
  begin
    p:=cross(q[h],q[h+1],q[r+1],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<=0 then inc(h) else inc(r);
    ans:=max(ans,dis(q[h],q[r]));
  end;
  writeln(ans);
end.