poj2386 Lake Counting

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 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.

 

题目大意是有一个园子，园子中W表示积水，积水周围8块有一块有积水就说明是联通的

问你总共有几块积水块？

代码如下：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
int dis[][2] = {{0,1},{0,-1},{1,1},{1,-1},{1,0},{-1,0},{-1,1},{-1,-1}};

int n,m;
char map[102][102];

void dfs(int x, int y) {
    map[x][y] = '.';
    for(int i = 0; i < 8; i++) {
        int xt = x + dis[i][0];
        int yt = y + dis[i][1];
        if(xt >= 0 && yt >= 0 && xt < n && yt < m && map[xt][yt] == 'W') {
            dfs(xt,yt);
        }
    }
}
int main(int argc, char const *argv[])
{
    //freopen("input.txt","r",stdin);
    while(scanf("%d %d",&n,&m) != EOF) {
        for(int i = 0; i < n; i++) {
            scanf("%s",map[i]);
        }
        int cnt = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(map[i][j] == 'W') {
                    dfs(i,j);
                    cnt++;
                }
            }
        }
        printf("%d
", cnt);
    }
    return 0;
}

采用深度优先搜索，搜索过的W改成.   ，直到没有W为止

搜索过几次就有几块