I - DFS（依然是漫水填充）

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

这道题依然是DFS求连通块，首先找到@，然后将它的值改成.,再对这一块进行填充就行，每填充一点，计数器就加一

#include"iostream"
#include"cstring"
#include"cstdio"
using namespace std;

char a[101][101];
int book[101][101];
int n,m;
int ans;

void dfs(int x,int y,int z)
{
if(x<0||x>=n||y<0||y>=m)
        return;
if(book[x][y]==1||a[x][y]!='.')
        return;
book[x][y]=z;
ans++;
for(int i=-1;i<=1;i++)
for(int j=-1;j<=1;j++)
if(j==0||i==0) dfs(x+i,y+j,z);
}

int main()
{
    while(scanf("%d%d",&m,&n)==2&&n&&m)
    {
   //  memset(a,0,sizeof(a));
     for(int i=0;i<n;i++) scanf("%s",a[i]);
     memset(book,0,sizeof(book));
     ans=0;
     for(int j=0;j<n;j++)
     for(int k=0;k<m;k++)
     {
        if(a[j][k]=='@') {a[j][k]='.';dfs(j,k,1);}
     }
cout<<ans<<endl;
    }
    return 0;
}