Lake Counting
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 17917 | Accepted: 9069 |
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.
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.
* 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
深搜
附代码
#include <iostream> #include <cstring> using namespace std; int n,m,ans,book[1010][1010]; char a[1010][1010]; int dx[] = {-1,-1,-1,0,0,1,1,1}; int dy[] = {-1,0,1,-1,1,-1,0,1}; void dfs(int x,int y) { for(int i = 0;i<8;i++) { int xx = x + dx[i]; int yy = y + dy[i]; if(xx >= 0 && xx < n && yy >=0 && yy < m && a[xx][yy] == 'W' && book[xx][yy] == 0) { book[xx][yy] = 1; dfs(xx,yy); } } } int main() { while(cin >> n >> m) { if(n == 0 && m == 0) break; memset(book,0,sizeof(book)); ans = 0; for(int i = 0;i<n;i++) for(int j = 0;j<m;j++) cin >> a[i][j]; for(int i = 0;i<n;i++) { for(int j = 0;j<m;j++) { if(a[i][j] == 'W' && book[i][j] == 0) { ans++; dfs(i,j); } } } cout<<ans<<endl; } }