Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1402 Accepted Submission(s): 904
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
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)
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
45 59 6 13
char a[25][25];
int flag[25][25];
int sum;
void DFS(int x,int y)
{
if(a[x-1][y]=='.' && flag[x-1][y]==0)
{
sum++;
flag[x-1][y]=1;
DFS(x-1,y);
}
if(a[x][y-1]=='.' && flag[x][y-1]==0)
{
sum++;
flag[x][y-1]=1;
DFS(x,y-1);
}
if(a[x][y+1]=='.' && flag[x][y+1]==0)
{
sum++;
flag[x][y+1]=1;
DFS(x,y+1);
}
if(a[x+1][y]=='.' && flag[x+1][y]==0)
{
sum++;
flag[x+1][y]=1;
DFS(x+1,y);
}
}
int main()
{
int n,m,i,j;
while(scanf("%d%d",&m,&n),m+n)
{
getchar();
sum=0;
memset(flag,0,sizeof(flag));
memset(a,'#',sizeof(a));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
scanf("%c",&a[i][j]);
}
getchar();
}
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(a[i][j]=='@')
{
sum++;
flag[i][j]=1;
DFS(i,j);
i=n;
j=m;
}
}
}
printf("%d/n",sum);
}
return 0;
}