原題連結

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42735		Accepted: 21161

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

3

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

題意：

有一個大小為M*N的園子，雨後積起了水。八連通的積水被認為是連線在一起的。請求出園子裡總共有多少水窪？（八連通指的是下圖中相對W的*的部分）

限制條件：N，M<=100;

#include<iostream>
#include<cstring>
using namespace std;
char maze[100][101];
int cnt,N,M;
//只要臨近的滿足條件就一直搜下去，並將滿足條件的置為'.'，直到所有的臨近全為'.'時才返回;
void dfs(int x,int y)
{
	maze[x][y]='.';//將滿足條件的位置全部置為'.';
	//用這兩個for迴圈迴圈遍歷8個方向；
	for(int i=-1;i<=1;i++)
		for(int j=-1;j<=1;j++)
		{
			int nx=x+i;
			int ny=y+j;
			if(nx>=0&&nx<N&&ny>=0&&ny<M&&maze[nx][ny]=='W')
				dfs(nx,ny);
		}
	return ;
}
int main()
{
	while(cin>>N>>M&&N>=1&&N<=100&&M>=1&&M<=100)
	{
		cnt=0;
		/*在讀取字元陣列的時候有兩種方法，第一種如程式中所示，注意別忘了加getchar()；
		第二種：用兩個for迴圈實現；
		for(int i=0;i<N;i++)
			for(int j=0;j<M;j++)
				cin>>maze[i][j];
		*/
		getchar();
		for(int i=0;i<N;i++)
			gets(maze[i]);
		for(int i=0;i<N;i++)
			for(int j=0;j<M;j++)
			{
				if(maze[i][j]=='W')//找到第一個時'W'的位置；
				{
					cnt++;//進行dfs的次數就是滿足條件的次數；
					dfs(i,j);
				}
			}
		cout<<cnt<<endl;
	}
	return 0;
}

還有一種DFS：

#include<iostream>
#include<cstring>
using namespace std;
const int MAX=110;
char arry[MAX][MAX];
int vist[MAX][MAX];
int num=0;
void DFS(int x,int y)
{
    if(arry[x][y]=='.'||vist[x][y]||arry[x][y]==0)
        return;
    vist[x][y]=1;
    DFS(x-1,y-1);   DFS(x-1,y);     DFS(x-1,y+1);
    DFS(x,y-1);                     DFS(x,y+1);
    DFS(x+1,y-1);   DFS(x+1,y);     DFS(x+1,y+1);
}
int main()
{
    int N,M;
    cin>>N>>M;
    num=0;
    memset(arry,0,sizeof(arry));
    memset(vist,0,sizeof(vist));
    for(int i=1;i<=N;i++)
        for(int j=1;j<=M;j++)
            cin>>arry[i][j];
    for(int i=1;i<=N;++i)
        {
            for(int j=1;j<=M;++j)
                if(!vist[i][j]&&arry[i][j]=='W')
                {
                    ++num;
                    DFS(i,j);
                }
        }
    cout<<num<<endl;
    return 0;
}