POJ 2386 Lake Counting

最新推荐文章于 2022-01-25 21:03:13 发布

Moressette

最新推荐文章于 2022-01-25 21:03:13 发布

阅读量693

点赞数 1

CC 4.0 BY-SA版权

分类专栏： DFS 文章标签： ACM DFS

本文链接：https://blog.youkuaiyun.com/wangcong9614/article/details/47204439

DFS 专栏收录该内容

5 篇文章

订阅专栏

本文解析了一道经典的DFS算法题目——湖泊计数问题。通过输入一个由水('W')和干地('.')组成的N行M列的地图，使用深度优先搜索算法来找出地图中湖泊的数量。文章提供了完整的AC代码及详细的实现步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

此题和HDU 1241 Oil Deposites题意是一样的，区别仅仅是表示的字符不一样（一个数油田一个数水洼→_→）。（HDU 1241题目链接已附在结尾处）题意很简单，输入两个整数N和M，表示一个N行M列的一块地，由输入的'W'和'.'组成，W代表该位置是水，'.'代表该位置是干地。相连的W组成一块水洼(相连指8个方向相连，即上、下、左、右、左上、左下、右上、右下)，求出一共有多少个水洼。经典dfs，直接看代码吧~

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23890		Accepted: 12067

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.

题目传送门： POJ2386 Lake Counting

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char maze[105][105];
int N,M;
int sum=0;
void dfs(int x,int y)
{
    if(x<0 || x>=N || y<0 || y>=M || maze[x][y]!='W')
        return;
    else
    {
        maze[x][y]='.';
        dfs(x+1,y);
        dfs(x-1,y);
        dfs(x,y+1);
        dfs(x,y-1);
        dfs(x-1,y-1);
        dfs(x+1,y-1);
        dfs(x-1,y+1);
        dfs(x+1,y+1);
    }
}

int main()
{
    sum=0;
    scanf("%d%d",&N,&M);
    for(int i=0;i<N;i++)
        for(int j=0;j<M;j++)
            cin>>maze[i][j];
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<M;j++)
        {
            if(maze[i][j]=='W')
                {
                    dfs(i,j);
                    sum++;
                }
        }
    }
    printf("%d\n",sum);
    return 0;
}

附： HDU1241 Oil Deposits题解&代码 HDU1241题目传送门： HDU1241 Oil Deposits