题目描述

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.

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水('W') 或是旱地('.')。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。

输入输出格式

输入格式：

 

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.

第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是'W'或'.'，它们表示网格图中的一排。字符之间没有空格。

 

输出格式：

 

Line 1: The number of ponds in Farmer John's field.

一行：水坑的数量

 

输入输出样例

输入样例#1： 

10 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.

输出样例#1： 

3

说明

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

思路：模拟

#include
         
          #include
          
           #include
           
            #include
            
             using namespace std;int n,m,ans;char s[110];int map[110][110];int dx[8]={
       1,-1,0,0,1,-1,-1,1};int dy[8]={
       0,0,1,-1,1,1,-1,-1};void dfs(int x,int y){    for(int i=0;i<8;i++){        int cx=dx[i]+x;        int cy=dy[i]+y;        if(cx>=1&&cx<=n&&cy>=1&&cy<=m&&map[cx][cy]){            map[cx][cy]=0;            dfs(cx,cy);        }    }}int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++){        scanf("%s",s);        for(int j=0;j