HDU 1241 Oil Deposits DFS(深さ優先探索)とBFS(広さ優先探索)

Oil Deposits
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8274 Accepted Submission(s): 4860
 
Problem Description
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
 
 
Input
The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.
 
 
Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
 
 
Sample Input
1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5 ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0
 
 
Sample Output
0 1 2 2
 
方法1:DFS(深さ優先探索)
 

import java.io.*;

import java.util.*;

public class Main {

	public  char ch[][];

	public int fx[]={1,1,1,-1,-1,-1,0,0};

	public int fy[]={0,1,-1,0,1,-1,1,-1};

	public int m,n;

	public static void main(String[] args) {

		new Main().work();

	}

	public void work(){

		Scanner sc=new Scanner(new BufferedInputStream(System.in));

		while(sc.hasNext()){

			m=sc.nextInt();

			n=sc.nextInt();

			ch=new char[m][n];

			if(m==0)

				System.exit(0);

			for(int i=0;i<m;i++){

				String s=sc.next();

				ch[i]=s.toCharArray();

			}

			int toal=0;

			for(int i=0;i<m;i++){

				for(int j=0;j<n;j++){

					if(ch[i][j]=='@'){

						toal++;

						dfs(i,j);

					}

				}

			}

			System.out.println(toal);

		}

	}

	public void dfs(int x,int y){

		for(int i=0;i<8;i++){

			int px=x+fx[i];

			int py=y+fy[i];

			if(check(px,py)){

				ch[px][py]='*';

				dfs(px,py);

			}

		}

	}

	public boolean check(int px,int py){

		if(px<0||px>m-1||py<0||py>n-1||ch[px][py]!='@')

			return false;

		return true;

	}

}

 
メソッド2 BFS(広さ優先探索)
 

import java.io.*;

import java.util.*;

public class Main {

	public int m,n;

	public char ch[][];

	public Queue<Bnode> list=new LinkedList<Bnode>();

	public int fx[]={1,1,1,-1,-1,-1,0,0};

	public int fy[]={0,1,-1,0,1,-1,1,-1};

	public static void main(String[] args) {

		new Main().work();

	}

	public void work(){

		Scanner sc=new Scanner(new BufferedInputStream(System.in));

		while(sc.hasNext()){

			list.clear();

			m=sc.nextInt();

			n=sc.nextInt();

			if(m==0)

				System.exit(0);

			ch=new char[m][n];

			for(int i=0;i<m;i++){

				String s=sc.next();

				ch[i]=s.toCharArray();

			}

			int total=0;

			for(int i=0;i<m;i++){

				for(int j=0;j<n;j++){

					if(ch[i][j]=='@'){

						total++;

						Bnode bnode=new Bnode();

						bnode.x=i;

						bnode.y=j;

						list.add(bnode);

						

						BFS();

					}

				}

			}

			System.out.println(total);

		}

	}

	public void BFS(){

		while(!list.isEmpty()){

			Bnode bnode=list.poll();

			for(int i=0;i<8;i++){

				int px=bnode.x+fx[i];

				int py=bnode.y+fy[i];

				if(check(px,py)){

					ch[px][py]='*';

					Bnode t=new Bnode();

					t.x=px;

					t.y=py;

					list.add(t);

				}

			}

		}

	}

	public boolean check(int px,int py){

		if(px<0||px>m-1||py<0||py>n-1||ch[px][py]!='@')

			return false;

		return true;

	}

}

class Bnode {

	int x;

	int y;

}