poj 2386 Lake counting

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

Hint

OUTPUT DETAILS: 

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

Source

这代码是挑战程序设计竞赛里的一份例题，代码思路也是按照书上的写的

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int n,m;char st[102][102];int dir[8][2]= {{0,1},{0,-1},{1,0},{-1,0},{1,-1},{1,1},{-1,-1},{-1,1}};void dfs(int x,int y) {    st[x][y]='.';    for(int i=0; i<8; i++) {        int newx=x+dir[i][0];        int newy=y+dir[i][1];        if(newx>=0 && newx <n && newy>=0 && newy <m && st[newx][newy]=='W') {            dfs(newx,newy);        }    }    return ;}int main() {    scanf("%d%d",&n,&m);    int ans=0;    for(int i=0; i<n; i++)        for(int j=0; j<m; j++)            scanf(" %c",&st[i][j]);    for(int i=0; i<n; i++)        for(int j=0; j<m; j++)            if(st[i][j]=='W') {                dfs(i,j);                ans++;            }    printf("%d\n",ans);}