POJ lake counting（简单的dfs）

A - Lake Counting

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

Hint

OUTPUT DETAILS: 

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

解题思路：

就是说，给你一个图，让你找出图中的所有Ｗ组成的联通块有多少个，其实就是ｄｆｓ求联通块的最简单的应用，每次找到‘Ｗ’后，就把‘Ｗ’修改成为‘．’然后，总共进行了多少次大的ｄｆｓ，就说明有几个联通块了。

代码：

# include<cstdio># include<iostream># include<algorithm># include<cstring># include<string># include<cmath># include<queue># include<stack># include<set># include<map>using namespace std;# define inf 999999999# define MAX 123char grid[MAX][MAX];int book[MAX][MAX];int n,m;int cnt;int next[8][2] = {{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1},{0,1},{1,1}};void input(){    for ( int i = 0;i < n;i++ )    {        scanf("%s",grid[i]);    }}void dfs( int x,int y ){    grid[x][y] = '.';    for ( int i = 0;i < 8;i++ )    {        int n_x = x+next[i][0];        int n_y = y+next[i][1];        if ( n_x>=0&&n_x<n&&n_y>=0&&n_y<m&&book[n_x][n_y]==0&&grid[n_x][n_y]=='W' )        {            dfs( n_x,n_y );        }    }}void solve(){    for ( int i = 0;i < n;i++ )    {        for ( int j = 0;j < m;j++ )        {            if ( grid[i][j]=='W' )            {                dfs( i,j );                cnt++;            }        }    }}int main(void){    while ( cin>>n>>m )    {        input();        solve();        cout<<cnt<<endl;    }return 0;}