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
45
59
6
13
方法一 DFS(深度優先搜素)
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; } }
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; } }