Red and Black Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 17584 Accepted: 9279 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) 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 #include<iostream> using namespace std; #define maxvex 21 char c[maxvex][maxvex]; struct {int x,y;}d[4]={{1,0},{-1,0},{0,1},{0,-1}}; int ans,line,row,newx,newy; void find(int i,int j) { ans++; c[i][j]='#'; for(int k=0;k<4;k++) { newx=i+d[k].x; newy=j+d[k].y; if(newx>=0&&newx<line&&newy>=0 && newy<row&&c[newx][newy]=='.') find(newx,newy); } } int main() { int i,j,tag; while(scanf("%d%d",&row,&line)&&line&&row) { tag=1; ans=0; for(i=0;i<line;i++) cin>>c[i]; for(i=0;i<line;i++) { for(j=0;j<row;j++) if(c[i][j]=='@') { find(i,j); break; tag=0; } if(tag==0) break; } printf("%d\n",ans); } return 0; }