POJ2386——Lake Counting

时间：2021-03-26 22:57:11 阅读：30 评论：0 收藏：0 [点我收藏+]

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.

作者解释：就是求上、下、左、右、左上、左下、右上、右下的连通块。

作者分析：dfs深搜，思路见细胞问题

#include <iostream>
#include <cstring>
using namespace std;

int idx[101][101],m,n,ans = 0;
char a[101][101];

void dfs(int x,int y,int id){
    if (x < 0 || x >= m || y < 0 || y >= n) return;
    if (idx[x][y] > 0 || a[x][y] == ‘.‘) return;
    idx[x][y] = id;
    dfs(x - 1,y,id);
    dfs(x + 1,y,id);
    dfs(x,y + 1,id);
    dfs(x,y - 1,id);
    if (x <= m - 1 && y <= n - 1) dfs(x + 1,y + 1,id);
    if (x >= 1 && y >= 1)dfs(x - 1,y - 1,id);
    if (x > 0 || y > 0) dfs(x + 1,y - 1,id);
    if (x <= m - 1 && y <= n - 1)dfs(x - 1,y + 1,id);
}

int main(){
    memset(a,0,sizeof(a));
    memset(idx,0,sizeof(idx));
    cin >> m >> n;
    for (int i = 0;i < m;i++){
        cin >> a[i];
    }
    int cnt = 0;
    for (int i = 0;i < m;i++){
        for (int j = 0;j < n;j++){
            if (idx[i][j] == 0 && a[i][j] != ‘.‘){
                dfs(i,j,++cnt);
            }
        }
    }
    cout << cnt; 
    return 0;
}

POJ2386——Lake Counting

原文：https://www.cnblogs.com/linyiweiblog/p/14584267.html

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