POJ 2386 Lake Counting（bfs解法）-优快云博客

本文介绍了一种通过广度优先搜索(BFS)算法来解决在二维矩阵中寻找池塘数量的问题。具体而言，该文详细解释了如何遍历由字符 'W' 和 '.' 组成的矩阵，其中 'W' 代表水，'.' 代表干地，并通过BFS算法确定彼此相连的水域形成的独立池塘的数量。

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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.

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

Hint

OUTPUT DETAILS:

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

Source

USACO 2004 November

#include <cstdio>
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

#define ll long long
int n, m, sum;;
char position[100+8][100+8], mov[8][2] = {0,1, 1,0, 0,-1, -1,0, 1,1, 1,-1, -1,-1, -1,1};

struct coordinate
{
    int x, y;
};

void bfs(int a, int b)
{
    queue<coordinate>q;
    coordinate miao;
    miao.x = a;
    miao.y = b;
    q.push(miao);
    while(!q.empty())
    {
        coordinate fir;
        fir = q.front();
        coordinate next;
        q.pop();//在这里就要弹出，不然后面会插入很多东西
        for(int i = 0; i<8; i++)
        {
            next.x = fir.x+mov[i][0];
            next.y = fir.y+mov[i][1];
            if(next.x >= 0 && next.x<n && next.y >= 0 && next.y<m && position[next.x][next.y] == 'W')
            {
                position[next.x][next.y] = '.';
                q.push(next);
            }
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    getchar();
    sum = 0;
    for(int i = 0; i<n; i++)
    {
        for(int j = 0; j<m; j++)
        {
            scanf("%c", &position[i][j]);
        }
        getchar();
    }
    for(int  i = 0; i<n; i++)
        for(int j = 0; j<m; j++)
            if(position[i][j] == 'W')
            {
                position[i][j] = '.';
                bfs(i, j);
                sum++;
            }
    printf("%d\n", sum);
    return 0;
}