HDU 1312 Red and Black DFS(深度优先搜索) 和 BFS（广度优先搜索）

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6762 Accepted Submission(s): 4284

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#..@#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###...@...###.###..#.#....#.#..0 0

Sample Output

4559613

方法一  DFS（深度优先搜素）

import java.io.*;import java.util.*;public class Main {int M=22,w,h,sx,sy;char ch[][];int fx[]={1,-1,0,0};int fy[]={0,0,1,-1};int number;boolean boo[][]=new boolean[100][100];public static void main(String[] args) {new Main().work();}void work(){Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNext()){w=sc.nextInt();h=sc.nextInt();if(h==0&&w==0)System.exit(0);ch=new char[h][w];for(int i=0;i<h;i++){String s=sc.next();ch[i]=s.toCharArray();Arrays.fill(boo[i], false);}for(int i=0;i<h;i++){for(int j=0;j<w;j++){if(ch[i][j]=='@'){sx=i;sy=j;}}}number=1;boo[sx][sy]=true;DFS(sx,sy);System.out.println(number);}}void DFS(int sx,int sy){for(int i=0;i<4;i++){int px=sx+fx[i];int py=sy+fy[i];if(check(px,py)&&!boo[px][py]){number++;boo[px][py]=true;DFS(px,py);}}}boolean check(int px,int py){if(px<0||px>h-1||py<0||py>w-1||ch[px][py]!='.')return false;return true;}}

方法二  BFS( 广度优先搜索)

import java.io.*;import java.util.*;public class Main {Queue<Node> que = new LinkedList<Node>();boolean boo[][] = new boolean[100][100];char ch[][];int w, h;int fx[] = { 1, -1, 0, 0 };int fy[] = { 0, 0, 1, -1 };int number;public static void main(String[] args) {new Main().work();}void work() {Scanner sc = new Scanner(new BufferedInputStream(System.in));while (sc.hasNext()) {w = sc.nextInt();h = sc.nextInt();if(h==0&&w==0)System.exit(0);ch = new char[h][w];for (int i = 0; i < h; i++) {String s = sc.next();ch[i] = s.toCharArray();Arrays.fill(boo[i], false);}Node node = new Node();for (int i = 0; i < h; i++) {for (int j = 0; j < w; j++) {if (ch[i][j] == '@') {node.x = i;node.y = j;node.number = 1;}}}boo[node.x][node.y] = true;que.add(node);number = 1;BFS();System.out.println(number);}}void BFS() {while (!que.isEmpty()) {Node node = que.poll();for (int i = 0; i < 4; i++) {int px = node.x + fx[i];int py = node.y + fy[i];if (check(px, py) && !boo[px][py]) {number++;Node td = new Node();td.x = px;td.y = py;boo[px][py] = true;ch[px][py] = 'S';que.add(td);}}}}boolean check(int px, int py) {if (px < 0 || px > h - 1 || py < 0 || py > w - 1 || ch[px][py] != '.')return false;return true;}class Node {int x;int y;int number;}}