1751: [Usaco2005 qua]Lake Counting

1751: [Usaco2005 qua]Lake Counting

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 190  Solved: 150
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Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M * Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,
and one along the right side.

HINT

 

Source

Gold

 

题解：直接萌萌哒DFS秒之，经典的普及组难度基础题，水水哒

(Tip:38行的dfs(a1,a2)貌似只有这样写在本机才能对，提交也能A；很神奇的是如果直接写dfs(i,j)的话在本机就会出现带入的是(1,1)结果进去的是(2,2)QAQ，然后各种神奇跪OTL，更神奇的是这个在本机都跪成狗的程序居然submit之后也能A（QAQ），求神犇解释)

var
   i,j,k,l,m,n,a1,a2:longint;
   c1:char;
   a:array[0..200,0..200] of longint;
procedure dfs(x,y:longint);inline;
          begin
               a[x,y]:=0;
               if a[x-1,y-1]=1 then dfs(x-1,y-1);
               if a[x,y-1]=1 then dfs(x,y-1);
               if a[x+1,y-1]=1 then dfs(x+1,y-1);
               if a[x-1,y+1]=1 then dfs(x-1,y+1);
               if a[x,y+1]=1 then dfs(x,y+1);
               if a[x+1,y+1]=1 then dfs(x+1,y+1);
               if a[x-1,y]=1 then dfs(x-1,y);
               if a[x+1,y]=1 then dfs(x+1,y);
          end;
begin
     readln(n,m);
     fillchar(a,sizeof(a),0);
     for i:=1 to n do
         begin
              for j:=1 to m do
                  begin
                       read(c1);
                       case c1 of
                            'W':a[i,j]:=1;
                            '.':a[i,j]:=0;
                       end;
                  end;
              readln;
         end;
     l:=0;
     for i:=1 to n do
         for j:=1 to m do
             if a[i,j]=1 then
                begin
                     a1:=i;a2:=j;
                     inc(l);dfs(a1,a2);
                end;
     writeln(l);
     readln;
end.