Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14311 Accepted Submission(s): 8870
Total Submission(s): 14311 Accepted Submission(s): 8870
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.
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)
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#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char map[500][500]; int n,m,ans,vis[500][500]; void dfs(int x,int y) { if(x<0||x>=m||y<0||y>=n||map[x][y]=='#'||vis[x][y]) return ; vis[x][y]=1; ans++; dfs(x+1,y); dfs(x,y+1); dfs(x-1,y); dfs(x,y-1); } int main() { while(scanf("%d%d",&n,&m),n||m) { int ex,ey; memset(map,' ',sizeof(map)); for(int i=0;i<m;i++) { scanf("%s",&map[i]); for(int j=0;j<n;j++) { if(map[i][j]=='@') { ex=i;ey=j;map[i][j]='.'; } } } ans=0; memset(vis,0,sizeof(vis)); dfs(ex,ey); printf("%d ",ans); } return 0; }