Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 23095 | Accepted: 12467 |
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
Source
简答的模板，
代码如下；
#include<stdio.h>
#include<string.h>
char a[22][22];
int v[22][22];
int sum;
void dfs(int i,int j)
{
if(a[i-1][j]=='.'&&v[i-1][j]==0)
{
sum++;
v[i-1][j]=1;
dfs(i-1,j);
}
if(a[i][j-1]=='.'&&v[i][j-1]==0)
{
sum++;
v[i][j-1]=1;
dfs(i,j-1);
}
if(a[i][j+1]=='.'&&v[i][j+1]==0)
{
sum++;
v[i][j+1]=1;
dfs(i,j+1);
}
if(a[i+1][j]=='.'&&v[i+1][j]==0)
{
sum++;
v[i+1][j]=1;
dfs(i+1,j);
}
}
int main()
{
int n,m,i,j;
while(~scanf("%d%d",&n,&m),n||m)
{
getchar();
sum=0;
memset(v,0,sizeof(v));
memset(a,'#',sizeof(a));
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%c",&a[i][j]);
}
getchar();
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
if(a[i][j]=='@')
{
v[i][j]=1;
sum+=1;
dfs(i,j);
break;
}
}
}
printf("%d\n",sum);
}
return 0;
}