思路：可以用DFS求解。遍历这个二维数组，没发现一次未被发现的‘@’，便将其作为起点进行搜索。最后的答案，是这个遍历过程中发现了几次为被发现的‘@’
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner in = new Scanner( System.in );
while( true ) {
int rows, cols;
rows = in.nextInt(); // read the rows
cols = in.nextInt(); // read the cols
if( rows == 0 && cols == 0 ) { // the input end
break;
}
OilDeposits od = new OilDeposits( rows, cols );
od.readGrids( in ); // read the strings
System.out.println( od.getPocketsAmount() );
}
}
}
class OilDeposits{
private int rows, cols;
private char grids[][];
private boolean vis[][];
private final int dir[][] = { // the eight directions
{-1,-1}, {-1,0}, {-1,1}, {0,-1}, {0,1}, {1,-1}, {1,0}, {1,1}
};
private final char EMPTY = '*';
private final char EXIST = '@';
public OilDeposits( int rows, int cols ) {
this.rows = rows;
this.cols = cols;
this.grids = new char[ rows + 5 ][ cols + 5 ];
this.vis = new boolean[ rows ][ cols ];
for( int i=0; i<rows; i++ ) {
for( int j=0; j<cols; j++ ) {
this.vis[i][j] = false;
}
}
}
private boolean isValid( int row, int col ) { // whether it can visit this position
if( row < 0 || row >= this.rows || col < 0 || col >= this.cols || vis[row][col] || this.grids[row][col] == EMPTY ) {
return false;
}
return true;
}
public void readGrids( Scanner in ) {
String str;
for( int i=0; i<rows; i++ ) {
this.grids[i] = in.next().toCharArray();
}
}
public void Search( int row, int col ) { // search this grid and its adjacent grids
if( !this.isValid( row, col ) ) {
return ;
}
this.vis[row][col] = true;
for( int i=0; i<8; i ++ ) {
Search( row+dir[i][0], col+dir[i][1] );
}
}
public int getPocketsAmount() {
int counter = 0;
for( int i=0; i<this.rows; i++ ) {
for( int j=0; j<this.cols; j++ ) {
if( this.grids[i][j] == '@' && !this.vis[i][j] ) {
counter ++;
this.Search( i, j );
}
}
}
return counter;
}
}
其中之所以将Scanner的对象进行传递，是因为如果重新声明一个Scanner的对象，会使输入出错，错误的具体表现为一直等待输入，尽管已经在Console中输入了数据。猜测可能是未关闭上一个Scanner的对象导致的，所以这里将Scanner的对象进行传递。