POJ2386 Lake Counting(dfs)

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17917 Accepted: 9069

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

Sample Output

3

dfs代码如下:

#include<cstdio>using namespace std;const int MAX_N=100;const int MAX_M=100;int N,M;char field[MAX_N][MAX_M+1];void dfs(int x,int y){field[x][y]='.';for(int dx=-1;dx<=1;dx++){for(int dy=-1;dy<=1;dy++){int nx=x+dx,ny=y+dy;if(0<=nx&&nx<N&&0<=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++;}}}printf("%d\n",res);}int main(){while(~scanf("%d%d",&N,&M)){for(int i=0;i<N;i++){scanf("%s",field[i]);}solve();}return 0;}