POJ 2386--Lake Counting [dfs] 《挑战程序设计竞赛》2.1

题目链接

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.

Source

USACO 2004 November

题目大意:

有一个大小为 N×M 的园子,雨后积起了水。八连通的积水被认为是连接在一起的。请求出
园子里总共有多少水洼?(八连通指的是下图中相对 W 的+的部分)
+++
+W+
+++

题解:

用f[][] 记录被访问过的位置, 用dfs搜索

代码:

#include <iostream>#define MAX_N 110#define MAX_M 110using namespace std;char a[MAX_N][MAX_M];bool f[MAX_N][MAX_M];int N, M;int cnt = 0;int fx[] = {0, 1, 0, -1, 1, 1, -1, -1};int fy[] = {1, 0, -1, 0, 1, -1, 1, -1};void dfs(int x, int y) {    f[x][y] = true;    for (int i = 0; i < 8; i++) {    int tx = x + fx[i];    int ty = y + fy[i];    if (!f[tx][ty] && a[tx][ty] == 'W')        dfs(tx, ty);    }}int main() {    cin >> N >> M;    for (int i = 0; i < N; i++)     for (int j = 0; j < M; j++)         cin >> a[i][j];    for (int i = 0; i < N; i++) {    for (int j = 0; j < M; j++) {        if (!f[i][j] && a[i][j] == 'W') {        cnt++;        dfs(i, j);        }    }    }    cout << cnt << endl;    return 0;}