POJ-1979 Red and Black-走方砖

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 26092		Accepted: 14163

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

Source

Japan 2004 Domestic

DFS搜索。

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define MAXN 15
char ma[21][21];
int dir[4][2]= {{-1,0},{0,1},{1,0},{0,-1}};
int m,n;
int dfs(int x,int y)
{
    int cnt=0;
    for(int i=0; i<4; ++i)
    {
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(x>=0||y>=0||x>n>>m&&m!=0&&n!=0)
    {
        memset(ma,'#',sizeof(ma));
        int cnt=0;
        int i,j,sx,sy;
        for(i=0; i>ma[i][j];
                if(ma[i][j]=='@')
                    sx=i,sy=j;
            }
        cnt=dfs(sx,sy);
        cout<<++cnt<