[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.

题解

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;

int m, n;
char garden[105][105];

void dfs(int x, int y)
{
    //将当前点取消标记，避免重复查找
    garden[x][y] = '.';

    //遍历周围的八个点
    for (int dx = -1; dx <= 1; dx++){
        for (int dy = -1; dy <= 1;dy++){
            int nx = x + dx;
            int ny = y + dy;
            if(0<=nx && nx<n && 0<=ny && ny<m && garden[nx][ny]=='W')
                dfs(nx, ny);
        }
    }
}

int main(void)
{
    while(~scanf("%d%d", &n,&m)){
        getchar();  //吸收两数字后的换行符
        memset(garden, 0, sizeof(garden));

        for (int i = 0; i < n;i++){
            for (int j = 0; j < m;j++)
                scanf("%c", &garden[i][j]);
            getchar();  //吸收每次输入一行后的换行符
        }

        int sum = 0;
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                if (garden[i][j] == 'W'){
                    dfs(i, j);
                    sum++;
                }
            }
        }

        cout << sum << endl;
    }

    return 0;
}