POJ 2386 Lake Counting （简单深搜）

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;
    }
}