Маяки могут находится в 1 точке.
P.S.
 Если считать, что маяки не могут быть в одной точке, то тогда нужно решать паросочетанием (как я в начале и решал), что гораздо сложнее по реализации

The problem was clarified:
It is guaranteed that measurements are consistent, i.e. the solution always exists, but may be ambiguous.

Incorrect test #15 was deleted. In the same time the new tests were added. In total, 27 authors lost their AC.

Anyone knows what's the problem with 2nd test? any ideas, please!!

Considering my submissions, I concluded that in the 15-th test there is a radiobeacon for which its coordinates can't be integers between 1 and 200. If for such radiobeacons output "UNKNOWN", the solution gets AC. This does not fit with the statement.

In test #15 there are radiobeacons which don't have correct location on the plane (i.e. no one of points 1 <= x,y <= 200 is acceptable) but their signals are identified by checkpoints. In the problem statement we can find, that "N radiobeacons are located on the plane" and "we know ... that their coordinates are integers from 1 to 200". So, I believe that such test cases are incorrect.

This test helped me to find several bugs
2
0,0:1-1,2012-200
201,201:2012-1

1:1,1
2012:200,200

I'm sure I got WA because of reading string.
Please help me, thanks.
Here is my code to read string :
...
fscanf(stdin,"%d\n",&m) ;
   for(int i=0;i<m;i++) {
       char s[1000] ;
         int j = 0 , k = 0 ;
         fscanf(stdin,"%s",&s) ;
         while ( s[j] != ',' ) a[i].x = a[i].x * 10 + s[j++] - 48 ;
         j++ ;
         while ( s[j] != ':' ) a[i].y = a[i].y * 10 + s[j++] - 48 ;
         j++ ;
         while (j < strlen(s) ) {
               while ( s[j] != '-' ) b[i][k] = b[i][k] * 10 + s[j++] - 48 ;
               j++ ;
               while ( j < strlen(s) && s[j] != ',' ) c[i][k] = c[i][k] * 10 + s[j++] - 48 ;
               k++ ;
               if ( s[j] == '\n' ) { j++ ; break ; }
               j++ ;
         }
         deg[i] = k ;
     }

This problem is simple, but I choose way to work with geometry for a long time.
Hint: while finding coordinates of beacon you must use information that beacon`s coordinates are between 1 and 200.

New tests have been added. 78 authors have lost AC.

While reading the information of the checkpoints,I can use there coordinates and the distance between it and a certain beacon to determin the range of the beacon.And as more information was read,the range will become smaller and smaller.So after the procedure,if I am sure the beacon is at a point,I output it and it's coordinates,otherwise I output UNKNOWN.Of course after sorting.

I think my algorithm is right if I did understand the idea.
But I got wa even on test 2,who can help me and tell me the meaning of the problem or give me your algorithm,thank you.

Why sample output isn't for example:

5:UNKNOWN
16:UNKNOWN

or

5:12,12
16:8,8?

Const MaxN      = 10;
      MaxSize   = 200;

Type TotalType  = Array[1..MaxN,1..MaxSize,1..MaxSize] of Byte;

Var  B          : ^TotalType;
     Id         : Array[1..MaxN] of Integer;
     T          : Array[1..MaxN] of Integer;
     N          : Integer;

Procedure FindPlace(X,Y,TheId,d : Integer);
Var i,j                         : Integer;
  Function GetNum : Integer;
    Var k         : Integer;
    Begin
      For k := 1 to N do
        If TheId=Id[k] then
          Begin
            GetNum := k;
            Exit;
          End;
      Inc(N); Id[N] := TheId;
      GetNum := N;
    End;
Begin
  i := GetNum; Inc(T[i]);
  For j := -d to d do
    Begin
      If (1<=X+d) and (X+d<=MaxSize) then
        If (1<=Y+j) and (Y+j<=MaxSize) then
          Inc(B^[i,X+d,Y+j]);
      If (1<=X-d) and (X-d<=MaxSize) then
        If (1<=Y+j) and (Y+j<=MaxSize) then
          Inc(B^[i,X-d,Y+j]);
    End;
  For j := -d+1 to d-1 do
    Begin
      If (1<=Y+d) and (Y+d<=MaxSize) then
        If (1<=X+j) and (X+j<=MaxSize) then
          Inc(B^[i,X+j,Y+d]);
      If (1<=Y-d) and (Y-d<=MaxSize) then
        If (1<=X+j) and (X+j<=MaxSize) then
          Inc(B^[i,X+j,Y-d]);
    End;
End;

Function ReadNum(Var X : String) : Integer;
Var Temp               : Longint;
Begin
  Temp := 0;
  While (Length(X)>0) and (X[1] in ['0'..'9']) do
    Begin
      Temp := Temp*10+Ord(X[1])-48;
      Delete(X,1,1);
    End;
  If Length(X)>0 then Delete(X,1,1);
  ReadNum := Temp;
End;

Procedure Main;
Var Total,i     : Integer;
    Str         : String;
    X,Y         : Integer;
    TheId,d     : Integer;
Begin
  Readln(Total);
  For i := 1 to Total do
    Begin
      Readln(Str);
      X := ReadNum(Str); Y := ReadNum(Str);
      While Length(Str)<>0 do
        Begin
          TheId := ReadNum(Str); d := ReadNum(Str);
          FindPlace(X,Y,TheId,d);
        End;
    End;
End;

Procedure Print;
Var i,j,k       : Integer;
    Total       : Integer;
    Place       : Array[1..MaxN,1..2] of Integer;
Begin
  Fillchar(Place,Sizeof(Place),0);
  For i := 1 to N do
    Begin
      Total := 0;
      For j := 1 to MaxSize do
        For k := 1 to MaxSize do
          If B^[i,j,k]=T[i] then
            Begin
              Inc(Total);
              Place[i,1] := j; Place[i,2] := k;
            End;
      If Total<>1 then
        Begin
          Place[i,1] := 0; Place[i,2] := 0;
        End;
    End;
  For j := 1 to 30000 do
    For i := 1 to N do
      If Id[i]=j then
        If Place[i,1]=0 then Writeln(j,':UNKNOWN')
                        else Writeln(j,':',Place[i,1],Chr(44),Place
[i,2]);
End;

Begin
  New(B);
  Fillchar(B^,Sizeof(B^),0);
    Main;
    Print;
  Dispose(B);
End.