【洛谷 P1596】[USACO10OCT] Lake Counting S 题解（深度优先搜索）

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[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.

样例输出 #1

3

提示

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

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思路

使用一个二维字符数组 gnd 存储地图信息。在搜索时，遍历整个地图，对于每个为 ‘W’ 的点，进行 DFS 搜索。在搜索时，将相邻的所有为 ‘W’ 的点都标记为 ‘.’，表示已经搜索过，避免重复搜索。在搜索结束后，计数器 sum 加一，表示搜索到一个湖泊。

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AC代码

#include <iostream>
#define AUTHOR "HEX9CF"
using namespace std;int n, m;
char gnd[100][100];
int dfs(int x, int y);
int sum = 0;int main()
{ios::sync_with_stdio(false);cin >> n >> m;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){cin >> gnd[i][j];// cout << gnd[i][j];}// cout << endl;}for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){if (gnd[i][j] == 'W'){dfs(i, j);sum++;}}}cout << sum << endl;return 0;
}int dfs(int x, int y)
{for (int i = x - 1; i <= x + 1; i++){for (int j = y - 1; j <= y + 1; j++){// cout << i << ',' << j << endl;if (gnd[i][j] == 'W' && i >= 0 && i <= n - 1 && j >= 0 && j <= m - 1){gnd[i][j] = '.';dfs(i, j);}}}return 0;
}

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