"Codevs" 1293

Input/Output sample

Idea: See the topic I Meng first thought of certainly is to beg unicom fast, but this unicom fast a bit peculiar, because

So he is a unicom fast. There are three ways to solve the problem: 1) wide search (this conforms to the Basic Law), 2) and check set; 3) Irrigation method

But Konjac Konjac wrote and looked at the set.

The code is as follows:

1 varN,m,i,j,ans:longint;2C:Array[0.. +,0.. +] ofChar;3PreArray[0..10000000] ofLongint;4A:Array[0..10000000] ofBoolean;5 functionFind (X:longint): Longint;6 begin7  ifPre[x]=x Thenexit (x);8find:=find (Pre[x]);9pre[x]:=find;Ten End; One procedurejoin (x,y:longint); A varFx,fy:longint; - begin -Fx:=find (x); fy:=find (y); the  ifFx<>fy Thenpre[fx]:=find (FY); - End; - begin - readln (n,m); +   fori:=1  toN*m Dopre[i]:=i; -   fori:=1  toN Do +   begin A     forj:=1  toM Do at     begin - read (c[i,j]); -          ifc[i,j]='#'  Thena[(I-1) *m+j]:=true; -          ifc[i,j]='-'  Thena[(I-1) *m+j]:=false; -     End; - Readln; in   End; -    fori:=1  toN Do to    begin +      forj:=1  toM Do -      begin the       ifc[i,j]='#'  Then *       begin $            ifc[i+1, j]='#'  ThenJoin (I*m+j, (i-1) *m+j);Panax Notoginseng            ifc[i,j+1]='#'  ThenJoin ((I-1) *m+j+1, (I-1) *m+j); -             ifc[i+2, j]='#'  ThenJoin ((i+1) *m+j, (I-1) *m+j); the             ifc[i,j+2]='#'  ThenJoin ((I-1) *m+j+2, (I-1) *m+j); +             ifc[i+1, J-1]='#'  ThenJoin (i*m+j-1, (I-1) *m+j); A             ifc[i+1, j+1]='#'  ThenJoin (i*m+j+1, (I-1) *m+j);//In fact, not 12 points all to be sentenced again, because there are repeated coverage of the part the       End; +      End; -    End; $    fori:=1  toN*m Do ifPre[i]=i Then ifA[i] ThenInc (ANS);//statistics "Unicom fast" number $ writeln (ans); - End.

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This is Konjac Konjac first ac of mention, send an essay commemorative (^_^)

"Codevs" 1293