P1596 [USACO10OCT]Lake Counting S——dfs连通块

[USACO10OCT]Lake Counting S

题面翻译

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 $N\times M(1\le N\le 100,1\le M\le 100)$ 的网格图表示。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。

输出一行，表示水坑的数量。

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.

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.

输出格式

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

样例 #1

样例输入 #1

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
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样例输出 #1

3
1

提示

OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.

分析

此题就是求连通块数，用dfs每从一个起点搜索结束后，确定一个连通块，然后ans++；在dfs中，此题是有8个方向可搜索，然后把走过的点标记；

#include

using namespace std;

int dx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};

char a[105][105];
int n, m, ans;

void dfs(int x, int y) {
    for (int i = 0; i < 8; i++) {
        int xx = x + dx[i];
        int yy = y + dy[i];
        if (xx >= 0 && yy >= 0 && xx < n && yy < m && a[xx][yy] == 'W') {
            a[xx][yy] = '.';
            dfs(xx, yy);
        }
    }
}

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 (a[i][j] == 'W') {
                dfs(i, j);
                ans++;
            }
        }
    }
    cout << ans;
    return 0;
}

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