Lake Counting POJ

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

Hint

OUTPUT DETAILS:

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

Code

/*^....0^ .1 ^1^..     011.^     1.0^ 1  ^    ^0.11 ^        ^..^0.           ^ 0^.0            1 .^.1             ^0 .........001^.1               1. .111100....01^00             ^   11^        ^1. .1^1.^                              ^0  0^.^                                 ^0..1.1                                   1..^1 .0                                     ^  ^00.                                     ^^0.^^ 0                                     ^^110.^0   0 ^                                     ^^^10.01^^     10  1 1                                      ^^^1110.101     10  1.1                                      ^^^1111110010    01  ^^                                        ^^^1111^1.^           ^^^10  10^ 0^ 1                                            ^^111^^^0.1^       1....^11     0                                               ^^11^^^ 0..  ....1^   ^ ^1.     0^                                               ^11^^^ ^ 1 111^     ^ 0.10   00 11                                               ^^^^^   1 0           1.0^  ^0  ^0                                                ^^^^    0            0.0^  1.0  .^                                               ^^^^    1 1          .0^.^  ^^  0^                             ^1                ^^^^     0.         ^.11 ^      11                             1.                ^^^     ^ ^        ..^^..^      ^1                             ^.^               ^^^       .0       ^.00..^      ^0                              01               ^^^       ..      0..^1 ..        .1                             ^.^              ^^^       1 ^  ^0001^  1.        00                              0.             ^^^        ^.0 ^.1. 0^.        ^.^                             ^.^            ^^^         ..0.01 .^^.         .^                  1001        ^^            ^^^         . 1^. ^ ^.         11                0.    1         ^           ^^          0.0  ^.          0              ^0       1                   ^^^          0.0.^  1.          0^             0       .1                   ^^^          ...1   1.          00            .        .1                  ^^^           ..1      1.         ^.           0         .^                  ^^            ..0.     1.          .^          .         0                                  ..1     1.          01          .        .                                 ^ 0^.^     00          ^0          1.       ^                                 1 1.0      00           .            ^^^^^^                                   ..^      00           01                                                    ..1.       00           10                                                   1 ^^.1       00           ^.                                            ^^^    .1..        00            .1                                        1..01    ..1.1         00           1.                                       ..^      10^ 1^         00           ^.1                                      0 1      1.1           00            00                                       ^  1   ^.           00            ^.^                                        10^  ^^1.1           00             00                                              10^..^           1.             ^.                                               1.0 1            ^.              00                 00                            .^^            ^.              ^ 1                00   ^0000^     ^               011 0             ^.               00.0^              ^00000   1.00.1              11. 1              0               1^^0.01                      ^^^                01.^              ^                1   1^^                                       ^.^1 1                                                                              0...                                                                              1 ^1                                                                               1^ ^                                                                             .01                                                                             ^ 1..                                                          1.1            ^0.0^ 0                                                           1..01^^100000..0^1 1                                                            ^ 1 ^^1111^ ^^0 ^                                                             ^ 1      1000^.1                                                               ^.^     .   00..                                                                1.1    0.   01.                                                                  .    1.   .^1.                                                                 1    1.   ^0^ .                                                                 ^.1 00    01^.0                                                                  001.     .^*/
// Virtual_Judge —— Lake Counting POJ - 2386.cpp created by VB_KoKing on 2019-05-05:15.
/* Procedural objectives：Variables required by the program:Procedural thinking：
从任意的W开始， 不停地把邻接的部分用.代替。1次DFS后与初始的这个W连接的所有W就都被替换成了.，因此直到图中不在存在W位置，总共进行DFS的次数就是答案。8个方向对应了8种状态转移，每个格子作为DFS的参数至多被调用一次，所以复杂度为O（8*N*M）。Functions required by the program:*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."FIGHTING FOR OUR FUTURE！！！
*/
#include <iostream>
using namespace std;int N,M;
char field[107][107];void dfs(int x,int y)
{field[x][y]='.';for (int dx = -1; dx < 2; dx++){for (int dy = -1; dy < 2; dy++){int nx=x+dx,ny=y+dy;if (-1<nx&&nx<N+1&&-1<ny&&ny<M&&field[nx][ny]=='W')dfs(nx,ny);}}return;
}void solve()
{int res=0;for (int i = 0; i < N; i++) {for (int j = 0; j < M; j++) {if (field[i][j]=='W'){dfs(i,j);res++;}}}cout<<res<<endl;
}int main()
{cin>>N>>M;for (int i = 0; i < N; i++)for (int j = 0; j < M; j++)cin>>field[i][j];solve();return 0;
}