Lake Counting
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14776 | Accepted: 7496 |
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 12W........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.
Source
MYCode:
#include<iostream>
#include<cstring> #include<cstdio> #define MAX 110 char map[MAX][MAX]; bool vis[MAX][MAX]; int dirt[8][2]={ {1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}}; using namespace std; void dfs(int x,int y) { vis[x][y]=1; int i; int tx,ty; for(i=0;i<8;i++) { tx=x+dirt[i][0]; ty=y+dirt[i][1]; if(!vis[tx][ty] && map[tx][ty]=='W') { dfs(tx,ty); } } }int main()
{ int n,m; while(scanf("%d%d",&n,&m)!=EOF) { int i,j; memset(vis,0,sizeof(vis)); for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { cin>>map[i][j]; } } int ans=0; for(i=1;i<=n;i++) { for(j=1;j<=m;j++) { if(map[i][j]=='W'&&!vis[i][j]) { ans++; dfs(i,j); } } } printf("%d\n",ans); } }//
DFS