POJ2386水洼问题优先深度搜索

最新推荐文章于 2024-07-04 22:55:56 发布

xgogoforit

最新推荐文章于 2024-07-04 22:55:56 发布

阅读量1.2k

点赞数

CC 4.0 BY-SA版权

分类专栏： poj 优先深度搜索

本文链接：https://blog.youkuaiyun.com/xgogoforit/article/details/50863937

poj 同时被 2 个专栏收录

10 篇文章

订阅专栏

优先深度搜索

4 篇文章

订阅专栏

本文介绍了一种使用深度搜索算法解决田地水洼问题的方法，通过判断相邻方格的状态来确定水洼的存在，并计算出田地中水洼的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目连接

http://poj.org/problem?id=2386

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

题意

…
.W.
…
解释：在一个W的附近八个点内有另外一个W，两个连成一个水洼。

给出一个图，找出里面最多有多少个水洼相连接

题解

优先深度搜索

代码

#include<iostream>
  #include<cstdio>
  #include<cstring>
  #include<string>
  #include<cmath>
  #include<string>
 #include<cctype>
  #include<algorithm>
  #include<vector>
#define INF  1e9+7
using namespace std;
int n,m;
char a[110][110];
void dfs(int q,int p){
    a[q][p]='.';
    for (int i = -1; i <=1 ; ++i)
    {
        for (int j = -1; j <=1 ; ++j)
        {
            if (a[q+i][p+j]=='W'&&q+i>=0&&q+i<n&&p+j>=0&&p+j<m)
            {   
                a[q+i][p+j]='.';
                dfs(q+i,p+j);
            }
        }
    }
}
int main(int argc, char const *argv[])
{
    scanf("%d%d",&n,&m);
    memset(a,0,sizeof(a));
//  fflush(stdin);   //清空输入流  上一个scanf预留的\n 
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < m; ++j)
        {
            cin>>a[i][j];
           // scanf("%c",&a[i][j]);
        }
    }
    int ans=0;
    for (int i = 0; i < n; ++i)
    {
        for (int j = 0; j < m; ++j)
        {
            if (a[i][j]=='W')
            {
                dfs(i,j);
                ans++;  
            }
        }
    }
    printf("%d\n",ans );
    return 0;
}