Red and Black（DFS）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1312

Problem：

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9944    Accepted Submission(s): 6190

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

dfs   code：

#include
#include
char map[21][21];
int vis[21][21];
int d[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
int w,h,cnt,sx,sy;
void dfs(int x,int y)
{
int i,xi,yi;
cnt++;
vis[x][y]=1;
for(i=0;i<4;i++)
{
xi=d[i][0]+x;
yi=d[i][1]+y;
if(xi>=0&&xi=0&&yi{

dfs(xi,yi);
}
}
}
int main()
{
int i,j;
while(scanf("%d%d",&w,&h)!=EOF)
{
if(w==0&&h==0)
break;
for(i=0;i{
scanf("%s",map[i]);
for(j=0;j{
if(map[i][j]=='@')
{
sx=i;
sy=j;
}
}
}
cnt=0;
memset(vis,0,sizeof(vis));
dfs(sx,sy);
printf("%d\n",cnt);
}
return 0;
}

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bfs  code：