Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 33405 | Accepted: 18176 |
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.
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)
The end of the input is indicated by a line consisting of two zeros.
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)
The end of the input is indicated by a line consisting of two zeros.
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
Source
Japan 2004 Domestic
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 char map[25][25]; 6 bool vis[25][25]; 7 int n,m,sx,sy,sum; 8 int dx[]={0,0,1,-1}; 9 int dy[]={1,-1,0,0}; 10 void Dfs(int x,int y){ 11 sum++;vis[x][y]=true; 12 for(int i=0;i<4;i++){ 13 int nx=x+dx[i],ny=y+dy[i]; 14 if(nx<=n&&nx>=1&&ny<=m&&ny>=1&& 15 map[nx][ny]=='.'&&!vis[nx][ny]) 16 Dfs(nx,ny); 17 } 18 } 19 void init(){ 20 memset(vis,false,sizeof(vis)); 21 memset(map,'0',sizeof(map)); 22 sum=0; 23 } 24 int main() 25 { 26 while(1){ 27 scanf("%d%d",&m,&n); 28 if(n==0&&m==0) break; 29 init(); 30 for(int i=1;i<=n;i++) 31 for(int j=1;j<=m;j++){ 32 cin>>map[i][j]; 33 if(map[i][j]=='@') sx=i,sy=j; 34 } 35 Dfs(sx,sy); 36 printf("%d ",sum); 37 } 38 return 0; 39 }