HDU1312 Red and Black(DFS)

C - Red and Black
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status

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

 
注意这个题只有四个方向了!上下左右，斜方向行不通。

#include<stdio.h>
#include<string.h>
int n,m,ans;
char map[21][21];
int vis[21][21];
int dir[4][2]= {0,1,1,0,0,-1,-1,0};
void dfs(int i,int j)
{
for(int k=0; k<4; k++)
{
int di=i+dir[k][0];
int dj=j+dir[k][1];
if(di>=0&&di<n&&dj>=0&&dj<m&&map[di][dj]=='.'&&vis[di][dj]==0)
{
ans++;
vis[di][dj]=1;
dfs(di,dj);
}
}
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
if(n==0&&m==0)
break;
else
{
int k,l;
memset(vis,0,sizeof(vis));
for(int i=0; i<n; i++)
scanf("%s",map[i]);
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
{
if(map[i][j]=='@')
{
k=i;
l=j;
vis[k][l]=1;
}
}
ans=1;
dfs(k,l);
printf("%d\n",ans);
}
}
return 0;
}