2338: [HNOI2011]数矩形 - BZOJ

因为已经看了一眼题解，知道是算中点和长度，相同时构成一个矩形，所以就把所有的线段算出来，然后排序，相同的就更新答案

为了避免误差，我们都用整数存，中点直接相加就行了，没必要除2，长度也只要平方就行了，不用开根，算面积就用叉积算，这样做就完全没有误差了

虽然复杂度本来是不行的但是出题人没想卡咱，就过了

 const
     maxn=;
 type
     point=record
       x,y:longint;
     end;
     segment=record
       aa,bb:point;
       x,y:longint;
       l:int64;
     end;
 var
     a:array[..maxn]of point;
     seg:array[..maxn*maxn]of segment;
     n,tot:longint;
     ans:int64;

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

 procedure sort(l,r:longint);
 var
     i,j:longint;
     x:segment;
 begin
     i:=l;
     j:=r;
     x:=seg[(l+r)>>];
     repeat
       while (x.x<seg[i].x) or ((x.x=seg[i].x) and (x.y<seg[i].y)) or ((x.x=seg[i].x) and (x.y=seg[i].y) and (x.l<seg[i].l)) do
         inc(i);
       while (x.x>seg[j].x) or ((x.x=seg[j].x) and (x.y>seg[j].y)) or ((x.x=seg[j].x) and (x.y=seg[j].y) and (x.l>seg[j].l)) do
         dec(j);
       if i<=j then
       begin
         swap(seg[i],seg[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,j:longint;
 begin
     read(n);
     for i:= to n do
       with a[i] do
       read(x,y);
     for i:= to n- do
       for j:=i+ to n do
         begin
           inc(tot);
           with seg[tot] do
             begin
               aa:=a[i];
               bb:=a[j];
               x:=a[i].x+a[j].x;
               y:=a[i].y+a[j].y;
               l:=sqr(int64(a[i].x-a[j].x))+sqr(int64(a[i].y-a[j].y));
             end;
         end;
     sort(,tot);
 end;

 function cj(a,b,c:point):int64;
 begin
     exit(abs(int64(a.x-b.x)*int64(c.y-b.y)-int64(c.x-b.x)*int64(a.y-b.y)));
 end;

 procedure up(var x:int64;y:int64);
 begin
     if x<y then x:=y;
 end;

 procedure work;
 var
     l,r,i,j:longint;
 begin
     l:=;
     r:=;
     while l<=tot do
       begin
         while (r<tot) and (seg[r+].x=seg[l].x) and (seg[r+].y=seg[l].y) and (seg[r+].l=seg[l].l) do
           inc(r);
         for i:=l to r- do
           for j:=i+ to r do
             up(ans,cj(seg[i].aa,seg[i].bb,seg[j].aa));
         l:=r+;
         r:=l;
       end;
     write(ans);
 end;

 begin
     init;
     work;
 end.