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.
题意:
在N*M的园子里下雨之后积水,’W’代表积水,’.’代表不积水,问积水的八连通区域有多少。(八连通是指一个格子周围的八个格子)
题解:
简单的dfs问题,从起点开始搜索是’W’的格子,搜到之后开始向八个方向搜索其他’W’的格子,一直搜索到没有连通的’W’格子,并标记为已经搜索过的,本题中我把值改为’.’,这样搜到多少次’W’,就有多少个连通区域。
#include <iostream>
#define MAXN 100+10
using namespace std;
const int dx[8] = {-1,-1,-1,0,0,1,1,1};
const int dy[8] = {1,0,-1,1,-1,1,0,-1};
char c[MAXN][MAXN];
int n,m;
void dfs(int x, int y)
{
//将'W'改为'.',表示已经搜索过,很聪明的一种办法
c[x][y] = '.';
for(int i = 0; i < 8; i++)
{
int nx = x + dx[i];
int ny = y + dy[i];
if(c[nx][ny] == 'W' && nx >= 1 && nx <= n && ny >= 1 && ny <= m)
dfs(nx,ny);
}
}
int main()
{
while(cin >> n >> m)
{
int ans = 0;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
cin >> c[i][j];
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
if(c[i][j]=='W')
{
dfs(i,j);
ans++;
}
cout << ans << endl;
}
return 0;
}