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.
- Line 1: The number of ponds in Farmer John’s field.
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
HintOUTPUT 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^
.. 01
1.^ 1.0
^ 1 ^ ^0.1
1 ^ ^..^
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 ^^^1110.1
01 10 1.1 ^^^1111110
010 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. ^.1
1 ^ 11 1. ^^^ ^ ^ ..^
^..^ ^1 ^.^ ^^^ .0 ^.0
0..^ ^0 01 ^^^ .. 0..^
1 .. .1 ^.^ ^^^ 1 ^ ^0001
^ 1. 00 0. ^^^ ^.0 ^.1
. 0^. ^.^ ^.^ ^^^ ..0.0
1 .^^. .^ 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^ ^ 01
1 0 ^. 00.0^ ^00000 1.00.1 11
. 1 0 1^^0.01 ^^^ 01
.^ ^ 1 1^^ ^.^
1 1 0.
.. 1 ^
1 1
^ ^ .0
1 ^ 1
.. 1.1 ^0.0
^ 0 1..01^^100000..0^
1 1 ^ 1 ^^1111^ ^^
0 ^ ^ 1 1000^
.1 ^.^ . 00
.. 1.1 0. 0
1. . 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 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; }
