Lake Counting
Time Limit: 1000MS | Memory Limit: 65536K |
Total Submissions: 26822 | Accepted: 13466 |
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
就是循環檢測發現是W 然後就深度搜尋旁邊的8個方塊,這裡用java ac了一發
import java.util.Scanner;
public class Main {
static int maxn = 105 ;
static char mapp[][] = new char[maxn][maxn];
static int n,m,ans;
static int addx []= {-1,-1,-1,0,0,1,1,1} ;
static int addy []= {-1,0,1,-1,1,-1,0,1} ;
static void dfs(int x,int y){
mapp[x][y]='.' ;
for(int i=0;i<8;i++){
int newx = x+addx[i] ;
int newy = y+addy[i] ;
if(newx>=0&&newx<=n&&newy>=0&&newy<=m&&mapp[newx][newy]=='W'){
//ans++ ;
dfs(newx,newy);
}
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in) ;
while(in.hasNextInt()){
n = in.nextInt() ;
m = in.nextInt() ;
for(int i=0;i<n;i++){
String str = in.next() ;
for(int j=0;j<m;j++){
mapp[i][j] = str.charAt(j) ;
}
}
ans = 0 ;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(mapp[i][j]=='W'){
ans++ ;
dfs(i,j);
}
}
}
System.out.println(ans);
}
}
}