poj2386 （DFS）

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28816 Accepted: 14437

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.

Source

USACO 2004 Novembe

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>char a[110][110];//标记园子int n,m;void DFS(int x,int y){    a[x][y]='.';//将搜索过的w的位置转换成'.'    for(int dx=-1;dx<=1;dx++)//循环遍历8个方向    {        for(int dy=-1;dy<=1;dy++)        {            int nx=dx+x;            int ny=dy+y;            if(nx>=0&&nx<n&&ny>=0&&ny<m&&a[nx][ny]=='W')                DFS(nx,ny);        }    }}int main(){    scanf("%d%d",&n,&m);    for(int i=0;i<n;i++)    {        scanf("%s",a[i]);    }    int sum=0;    for(int i=0;i<n;i++)    {        for(int j=0;j<m;j++)        {            if(a[i][j]=='W')//从W位置搜索            {                DFS(i,j);                sum++;            }        }    }    printf("%d\n",sum);}