题目描述

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.

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水(‘W’) 或是旱地(’.’)。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。
输入格式

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.

第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是’W’或’.’，它们表示网格图中的一排。字符之间没有空格。
输出格式

Line 1: The number of ponds in Farmer John’s field.

一行：水坑的数量

输入：

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

输出：

3

解题思路：
找到一个W，从这个W开始dfs，把8个方向的W都消除，然后继续找W，计算次数。

代码如下：

#include <iostream>
using namespace std;
const int N = 110;
int ans;int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};int dy[] = {1, -1, 0, 0, -1, -1, 1, 1};
char g[N][N];
int n, m;void dfs(int x, int y) {g[x][y] = '.';for (int i = 0; i < 8; i++) {int xx = x + dx[i], yy = y + dy[i];if (xx >= 0 && xx < n && yy >= 0 && yy < m && g[xx][yy] == 'W') {dfs(xx, yy);}}
}int main() {cin >> n >> m;for (int i = 0; i < n; i++)cin >> g[i];for (int i = 0; i < n; i++)for (int j = 0; j < m; j++) {if (g[i][j] == 'W') {dfs(i, j);ans++;}}cout << ans << endl;return 0;
}