Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8694 Accepted Submission(s): 5393
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
#include<stdio.h>
#include<string.h>
char map[21][21];
int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int n,m,t,visit[21][21];
void dfs(int xx,int yy)
{
int x,y,l;
for(l=0;l<4;l++)
{
x=xx+dx[l];
y=yy+dy[l];
if(x>=0&&x<n&&y>=0&&y<m&&map[x][y]=='.'&&visit[x][y]==0)
{
visit[x][y]=1;
t++;
dfs(x,y);
}
}
}
main()
{
int i,j;
while(scanf("%d%d",&m,&n)&&!(m==0&&n==0))
{
getchar();
memset(visit,0,sizeof(visit));
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
scanf("%c",&map[i][j]);
getchar();
}
for(i=0;i<n;i++)
for(j=0;j<m;j++)
if(map[i][j]=='@')
{
t=1;
visit[i][j]=1;
dfs(i,j);
}
printf("%d\n",t);
}
}