Lake Counting（深度优先算法）-优快云博客

本文介绍了一种通过深度优先遍历算法来计算二维矩阵中八连通水洼数量的方法，并提供了两种实现方式。

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这个题的模板也可以AC 另两个题，代码意思都是差不多的

Red and Black（找黑砖的个数）点击打开链接

Oil Deposits（计算油田的个数）点击打开链接

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.

大致题意

有一个大小为N*M的园子，雨后积了很多水。八连通的积水被认为是在一起的。请求出园子里共有多少个水洼？(八连通是指下图中相对.的8 部分)

www

w.w

www

从任意的'W'开始，不停地把邻接的部分用'.'代替，一次DFS(深度优先遍历)遍历后，与初始的这个 W 所连接的所有 'W' 都会被替换成 '.'，因此直到图中没有 'W'为止，总共进行 DFS 的次数即为积水的次数。

代码如下：

第一种方法；

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char a[1001][1001];
int n,m;
void DFS(int x,int y)
{
    int i,j,b,c;
    for(i=-1;i<=1;i++){
        for(j=-1;j<=1;j++){
            b=i+x;//代表了八个方向  (i-1,j-1)  (i-1,j)  (i-1,j+1) (i,j-1)
            c=y+j;//		  (i,j+1)   (i+1,j-1)  (i+1,j)  (i+1,j+1)
            if(b>=0&&b<n&&y>=0&&y<m&&a[b][c]=='W')
            {
                a[b][c]='.';//如果改点是 ’W' 的话 就另 该点是 '.' 以后也不访问他
                DFS(b,c);// 去搜索该点
            }
        }
    }
}
int main()
{
    int i,j,flag=0;
    while(scanf("%d %d",&n,&m)!=EOF){
        memset(a,'0',sizeof(a));
        flag=0;
        for(i=0;i<=n-1;i++)
            scanf("%s",a[i]);
        for(i=0;i<=n-1;i++){
            for(j=0;j<=m-1;j++){
                if(a[i][j]=='W'){
                    flag++;//遇见一个水洼 就加加
                    DFS(i,j);//去搜索这一点的八个方向
                }
            }
        }
        printf("%d\n",flag);
    }
}

第二种方法：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char a[1001][1001];
int n,m;
void DFS(int i,int j)
{
    if(i>=0&&i<n&&j>=0&&j<m&&a[i][j]=='W'){
        a[i][j]='.';//相比于第一种，这里用了八个 DFS，第一种用了循环来表示八个方向
        DFS(i-1,j-1);
        DFS(i-1,j);
        DFS(i-1,j+1);
        DFS(i,j-1);
        DFS(i,j+1);
        DFS(i+1,j-1);
        DFS(i+1,j);
        DFS(i+1,j+1);
    }
}
int main()
{
    int i,j,flag=0;
    while(scanf("%d %d",&n,&m)!=EOF){
        memset(a,'0',sizeof(a));
        flag=0;
        for(i=0;i<=n-1;i++)
            scanf("%s",a[i]);
        for(i=0;i<=n-1;i++){
            for(j=0;j<=m-1;j++){
                if(a[i][j]=='W'){
                    flag++;
                    DFS(i,j);
                }
            }
        }
        printf("%d\n",flag);
    }
}