[POJ2002 Squares]

[题目来源]：POJ2002

[关键字]：Hash （二分）

[题目大意]：给出n个点，求它们共组成了多少个正方形。

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[分析]：因为是正方形所以只要枚举任意两点即可求出另外两点的坐标，数学问题在此不再诉述可以自己推一推。现在的任务就是判断求出的可以构成正方形的两点是否存在，先排序这样枚举时可以节省时间。至于查找有两种方法：1、hash每一点坐标值（x^2+y^2 mod 素数），然后查找。2、二分查找。hash过了，二分总是超时........

[代码]：

View Code

  1 program Project1;
  2 type
  3   rec = record
  4     x, y: longint;
  5   end;
  6 var
  7   h: array[0..20000,0..20] of rec;
  8   a: array[0..2000] of rec;
  9   num: array[0..20000] of longint;
 10   n, ans: longint;
 11 
 12 procedure qs(l, r: longint);
 13 var
 14   i, j, midx, midy: longint;
 15   t: rec;
 16 begin
 17   i := l;
 18   j := r;
 19   midx := a[(l+r) shr 1].x;
 20   midy := a[(l+r) shr 1].y;
 21   repeat
 22     while (a[i].x < midx) or ((a[i].x = midx) and (a[i].y < midy)) do inc(i);
 23     while (a[j].x > midx) or ((a[j].x = midx) and (a[j].y > midy)) do dec(j);
 24     if i <= j then
 25       begin
 26         t := a[i];
 27         a[i] := a[j];
 28         a[j] := t;
 29         inc(i);
 30         dec(j);
 31       end;
 32   until i > j;
 33   if l < j then qs(l,j);
 34   if i < r then qs(i,r);
 35 end;
 36 
 37 function hash(x, y: longint):longint;
 38 begin
 39   hash := (x*x+y*y) mod 15988;
 40 end;
 41 
 42 procedure init;
 43 var
 44   i, k: longint;
 45 begin
 46   readln(n);
 47   if n = 0 then halt;
 48   fillchar(num,sizeof(num),0);
 49   fillchar(h,sizeof(h),0);
 50   for i := 1 to n do
 51     begin
 52       readln(a[i].x,a[i].y);
 53       k := hash(a[i].x,a[i].y);
 54       inc(num[k]);
 55       h[k,num[k]].x := a[i].x;
 56       h[k,num[k]].y := a[i].y;
 57     end;
 58   qs(1,n);
 59 end;
 60 
 61 function find(x, y: longint):boolean;
 62 var
 63   k, i: longint;
 64 begin
 65   k := hash(x,y);
 66   for i := 1 to num[k] do
 67     if (h[k,i].x = x) and (h[k,i].y = y) then exit(true);
 68   exit(false);
 69 end;
 70 
 71 procedure work;
 72 var
 73   i, j, x1, y1, x2, y2, tempx, tempy: longint;
 74   f1, f2: boolean;
 75 begin
 76   ans := 0;
 77   for i := 1 to n do
 78     for j := i+1 to n do
 79       begin
 80         tempx := a[j].x-a[i].x;
 81         tempy := a[j].y-a[i].y;
 82         x1 := a[i].x-tempy;
 83         y1 := a[i].y+tempx;
 84         x2 := a[j].x-tempy;
 85         y2 := a[j].y+tempx;
 86         //writeln(a[i].x,' ',a[i].y,' ',a[j].x,' ',a[j].y,' ',x1,' ',y1,' ',x2,' ',y2);
 87         f1 := find(x1,y1);
 88         f2 := find(x2,y2);
 89         if f1 and f2 then inc(ans);
 90         //writeln(ans);
 91       end;
 92   ans := ans shr 1;
 93   writeln(ans);
 94 end;
 95 
 96 begin
 97   while 1 = 1 do
 98     begin
 99       init;
100       work;
101     end;
102 end.