POJ - 2386 dfs-Lake Counting

最新推荐文章于 2022-01-25 21:03:13 发布

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最新推荐文章于 2022-01-25 21:03:13 发布

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分类专栏： dfs练习题文章标签： dfs

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本文介绍了一种使用深度优先搜索（DFS）算法解决农田积水连通区域计数问题的方法。通过遍历N*M矩阵，识别并计算出所有八连通积水区域的数量。

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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.

题意：

在N*M的园子里下雨之后积水，’W’代表积水，’.’代表不积水，问积水的八连通区域有多少。（八连通是指一个格子周围的八个格子）

题解：

简单的dfs问题，从起点开始搜索是’W’的格子，搜到之后开始向八个方向搜索其他’W’的格子，一直搜索到没有连通的’W’格子，并标记为已经搜索过的，本题中我把值改为’.’，这样搜到多少次’W’，就有多少个连通区域。

#include <iostream>
#define MAXN 100+10
using namespace std;

const int dx[8] = {-1,-1,-1,0,0,1,1,1};
const int dy[8] = {1,0,-1,1,-1,1,0,-1};
char c[MAXN][MAXN];
int n,m;

void dfs(int x, int y)
{
    //将'W'改为'.'，表示已经搜索过，很聪明的一种办法
    c[x][y] = '.';
    for(int i = 0; i < 8; i++)
    {
        int nx = x + dx[i];
        int ny = y + dy[i];
        if(c[nx][ny] == 'W' && nx >= 1 && nx <= n && ny >= 1 && ny <= m)
            dfs(nx,ny);
    }
}

int main()
{
    while(cin >> n >> m)
    {
        int ans = 0;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                cin >> c[i][j];
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                if(c[i][j]=='W')
                {
                    dfs(i,j);
                    ans++;
                }
        cout << ans << endl;
    }
    return 0;
}