Lake Counting DFS水题

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

#include <stdio.h>#define  max 102int dx[8]={-1,-1,-1,0,0,1,1,1};int dy[8]={-1,0,1,-1,1,-1,0,1};char map[max][max];int x[max*100],y[max*100];int n,m,num=0;void DFS(int s,int k){int x1,y1;int i;for(i=0;i<8;++i){x1=s+dx[i];y1=k+dy[i];if(x1>=1&&x1<=n&&y1>=1&&y1<=m&&map[x1][y1]=='W'){map[x1][y1]='.';DFS(x1,y1);}}return ;} int main(){//freopen("input.txt","r",stdin);int i,j,s=0,k=0;scanf("%d%d",&n,&m);for(i=1;i<=n;++i)for(j=1;j<=m;++j){scanf(" %c",&map[i][j]);if(map[i][j]=='W'){x[s++]=i;y[k++]=j;}}for(i=0;i<s;++i){if(map[x[i]][y[i]]=='W'){map[x[i]][y[i]]='.';DFS(x[i],y[i]);++num;}}printf("%d\n",num);return 0;}