Lake Counting

题目：

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.

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良心翻译：

描述

由于最近的降雨，水汇集在Farmer John’s田地的不同地方，其由N×M（1 <= N <= 100; 1 <= M <= 100）的正方形矩形表示。每个方块包含水（‘W’）或旱地（’.’）。农夫约翰想弄清楚他的田地里有多少个池塘。池塘是一组连接的正方形，其中有水，其中一个正方形被认为与其所有八个邻居相邻。

给出农夫约翰的田地图，确定他有多少池塘。
输入

*第1行：两个以空格分隔的整数：N和M.

*行2…N + 1：每行M个字符代表一行Farmer John的字段。每个字符都是’W’或’.’。字符之间没有空格。
产量

*第1行：Farmer John的田地里的池塘数量。

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思路：

题目一大段，只是让你做一件事情：数连通块，DFS一波就行了。

代码：

#include<bits/stdc++.h>
using namespace std;
int n,m;
int maze[1010][1010];
bool vis[1010][1010];
 
bool in(int x,int y){
    return x>=1&&x<=n&&y>=1&&y<=m;
}//防止出界
 
int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{-1,-1},{1,1},{-1,1},{1,-1}};//八个方向
void dfs(int sx,int sy){
    vis[sx][sy]=1;//标记起点
    for(int i=0;i<8;i++){
        int tx=sx+dir[i][0];
        int ty=sy+dir[i][1];
        if(!vis[tx][ty] && maze[tx][ty]==1&&in(tx,ty)){
            dfs(tx,ty);/继续dfs
        }
    }
    return;
}
int main(){
    int ans=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            char pi;
            cin>>pi;
            if(pi=='W'){
                maze[i][j]=1;
            }else{
                maze[i][j]=0;
            }
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(maze[i][j]==1 && !vis[i][j]){
                dfs(i,j);
                ans++;//枚举点，dfs它后，和他联通的块就被标记了，答案数+1
            }
        }
    }
    cout<<ans;
    return 0;
}