初级dfs

POJ NO.2386：https://vjudge.net/problem/POJ-2386
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 12
W........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 November
*/

题意：找八个方位的不连通块有多少个

 

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m;
char mp[106][106];
void DFS(int i,int j)
{
    int dx,dy,nx,ny;
    if(i<0||i>=m||j<0||j>=n)
        return ;
    mp[i][j]='.';
    for(dx=-1;dx<=1;dx++)
        for(dy=-1;dy<=1;dy++)
    {
        nx=i+dx;
        ny=j+dy;
    if(nx>=0&&nx<m&&ny>=0&&ny<n&&mp[nx][ny]=='W')DFS(nx,ny);
    }
    return ;
}
int main()
{
    int i,j,res=0;
        cin>>m>>n;
        getchar();
        for(i=0;i<m;i++)
        {

            for(j=0;j<n;j++)
        {
            scanf("%c",&mp[i][j]);
        }
                    getchar();
        }
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
        {
            if(mp[i][j]=='W')
            {
                DFS(i,j);
                res++;
            }
        }
        cout<<res<<endl;
    return 0;
}