湖计数

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<iostream>#include<string.h>#include<stdio.h>#include<ctype.h>#include<algorithm>#include<stack>#include<queue>#include<set>#include<math.h>#include<vector>#include<deque>#include<list>using namespace std;char map[120][120];int m,n;void dfs(int x,int y){    if(map[x][y]!='W'||x<0||y<0||x>=m||y>=n)        return;    else    {        map[x][y]='.';        dfs(x-1, y-1);        dfs(x-1, y);        dfs(x-1, y+1);        dfs(x, y-1);        dfs(x, y+1);        dfs(x+1, y-1);        dfs(x+1, y);        dfs(x+1, y+1);    }}int main(){    scanf("%d%d",&m,&n);    for(int i=0; i<m; i++)        for(int j=0; j<n; j++)            cin>>map[i][j];    int count=0;    for(int i=0; i<m; i++)        for(int j=0; j<n; j++)        {            if(map[i][j]=='W')            {                dfs(i,j);                count++;            }        }    printf("%d\n",count);    return 0;}