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
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13
59
6
13
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; }
}