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

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： 從任意的W開始， 不停地把鄰接的部分用.代替。1次DFS后與初始的這個W連接的所有W就都被替換成了.，因此直到圖中不在存在W位置，總共進行DFS的次數就是答案。8個方向對應了8種狀態(tài)轉移，每個格子作為DFS的參數至多被調用一次，所以復雜度為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; }

總結

以上是生活随笔為你收集整理的Lake Counting POJ - 2386的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。