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

[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 × M ( 1 ≤ N ≤ 100 , 1 ≤ M ≤ 100 ) N\times M(1\leq N\leq 100, 1\leq M\leq 100) N×M(1N100,1M100) 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 1 1 行:两个空格隔开的整数: N N N M M M

2 2 2 行到第 N + 1 N+1 N+1 行:每行 M M M 个字符,每个字符是 W.,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.


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。


AC代码

#include 
#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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