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
Sample Input W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W. Sample Output
题意：
W表示水洼；.表示空地，八个方向连起来的水洼也算是一个，问有多少水洼。
思路：
简单的搜索题目，这里用深搜实现。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
char map[][];//用来存储图的信息 /*标记是否搜索过，本题目没有使用，因为是一次访问，直接将访问过的修改即可*/
int logo[][]; int m,n,sum; /*表示八个方向，四个方向时，将后面四组删掉就可以了*/
int dir[][]= {,, ,-, ,, -,, ,, ,-, -,, -,-}; void DFS(int x,int y)
{
if(x>=&&y>=&&x<n&&y<m)//这里是判断是否越界，根据题目要求改写
{
if(map[x][y]=='W')//如果符合条件就继续递归。
{
map[x][y]='.';//标记为‘.’防止多次访问
for(int i=; i<; i++)//因为八个方向，所以循环八次。
DFS(x+dir[i][],y+dir[i][]);
}
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
sum=;
memset(logo,,sizeof(map));
getchar();
for(int i=; i<n; i++)
{
for(int j=; j<m; j++)
{
cin>>map[i][j];
}
getchar();
}
for(int i=; i<n; i++)
{
for(int j=; j<m; j++)
if(map[i][j]=='W')
{
DFS(i,j);
sum++;//计数
}
}
cout<<sum<<endl;
}
}