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lake counting -- DFS 搜索

搜索 -- DFS Counting Lake
2023-09-27 14:28:36 时间

问题

http://poj.org/problem?id=2386

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.

 

思路

对于每一个water ponds， 需要使用dfs 搜索其边界。

使用二维遍历， 当找到第一个为w的square，此点为一个ponds的部分， 从此点开始dfs搜索water ponds所有W块和边界。

 

 

Code

#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <deque>
#include <queue>
#include <string>
#include <set>
using namespace std;

/*
http://poj.org/problem?id=2386
*/

bool wfields[102][102] = {{false}};

bool visited[102][102] = {{false}};
int n, m;


void dfs(int i, int j) {
    if (i < 1 || i > n ) {
//        cout << "i is out of bound. i=" << i << endl;
        return;
    }

    if (j < 1 || j > m) {
//        cout << "j is out of bound. j=" << j << endl;
        return;
    }

    if (visited[i][j] == true) {
        return ;
    } else {
        visited[i][j] = true;
    }

    if (wfields[i][j] == false) {
//        cout << "water fields is false." << endl;
        return;
    }

    for (int diff_i = -1; diff_i <= 1; diff_i++) {
        for (int diff_j = -1; diff_j <= 1; diff_j++) {
            dfs(i + diff_i, j + diff_j);
        }
    }
}


int main() {
    cin >> n >> m;

    for (int i = 1; i <= n; i++) {
        string temp;
        cin >> temp;

        for (int j = 1; j <= m; j++) {
            if (temp[j - 1] == 'W') {
                wfields[i][j] = true;
            } else {
                wfields[i][j] = false;
            }

            visited[i][j] = false;
        }
    }

    int count = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (wfields[i][j] == true && visited[i][j] == false) {
                count ++;

                dfs(i, j);
            }
        }
    }

    cout << count << endl;
}

 

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