Lake Countinge的s

最新推荐文章于 2020-07-05 19:46:41 发布

原创最新推荐文章于 2020-07-05 19:46:41 发布 · 664 阅读

2 ·

CC 4.0 BY-SA版权

解题报告同时被 3 个专栏收录

71 篇文章

订阅专栏

====搜索====

11 篇文章

订阅专栏

DFS

6 篇文章

订阅专栏

本文介绍了一种使用深度优先搜索(DFS)算法来解决计算地图中湖泊数量的问题。地图由NxM的矩阵构成，每个格子代表水('W')或干地('.'). +

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

解题思路：

题目大意就是问你图中有几个湖泊，W周围八个方向中的W都归为一个湖泊，是连通的。DFS水题，连回溯都不需要，直接往八个方向检索下去，只要是W，就把＇W＇转变成＇．＇。目的是为了下次检索的时候不去访问之前的W。

AC代码：

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

const int maxn = 100 + 5;
char map[maxn][maxn];
int dir[8][2] = {{-1,-1},{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1}};  // 用二维数组储存八个方向

int dfs(int x,int y,int n,int m)
{
     int a, b;
     map[x][y] = '.';      // 讲访问过的"W"转化为"."
     for(int i = 0;i < 8; i++)
     {
          a = x + dir[i][0];
          b = y + dir[i][1];
          if(a < n && a >= 0 && b < m && b >= 0 && map[a][b] == 'W')
              dfs(a, b, n, m);
     }
     return 1;
}
int main()
{
    int n, m, ans = 0;
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i++)
        scanf("%s", map[i]);
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            if(map[i][j]=='W')       // 检索整个图
                ans += dfs(i,j,n,m);
    printf("%d\n", ans);
    return 0;
}