【练题】Lake Counting

最新推荐文章于 2024-11-06 08:22:20 发布

quchen528

最新推荐文章于 2024-11-06 08:22:20 发布

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本文介绍了一种使用深度优先搜索算法解决在二维矩阵中寻找池塘（连续水域）数量的问题。通过遍历矩阵，每当遇到水区域时进行深度优先搜索并将其标记为已访问，以此来计算总的池塘数量。

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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.

输入描述:

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.

输出描述:

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

示例1

输入

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

输出

解题思路

深搜每个点位，并把他 ‘W’ 填为 ‘.’，这样每遇一次水，计数一次，且将附近的所有水填掉，全局遍历解决，此题遍历每个点不是复杂度的关键，减少不必要的探索水关键，并非多好思路，您若有更好思路，敬请评论或私信。

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int n,m;
void Dfs(vector<string> &sv,int x,int y)
{
    if(x >= 0 && x < n && y >= 0 && y < m && sv[x][y] == 'W')
    {
        sv[x][y] = '.';
        for(int i = -1;i < 2;++i)
        {
            for(int j = -1;j < 2;++j)
            {
                if(!i && !j)
                    continue;
                Dfs(sv,x + i,y + j);
            }
        }
        
    }
}
int main()
{
    while(cin >> n >> m)
    {
        int count = 0;
        vector<string> sv(n);
        for(auto &e : sv)
            cin >> e;
        for(int i = 0;i < n;++i)
        {
            for(int j = 0;j < m;++j)
            {
                if(sv[i][j] == 'W')
                {
                    Dfs(sv,i,j);
                    ++count;
                }
            }
        }
        cout << count << endl;
    }
    return 0;
}