POJ练习深搜

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20512		Accepted: 10986

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

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<algorithm>
#define MAX 30
using namespace std;
char g[MAX][MAX];
int n, m, bx, by, totle;//行、列,起始坐标,总结果
int dx[4] = { -1, 1, 0, 0 }, dy[4] = { 0, 0, 1, -1 };
void dfs(int x, int y){
    g[x][y] = '#';
    for (int i = 0; i < 4; i++){
        int tx = x + dx[i], ty = y + dy[i];
        if (0 <= tx&&tx < n && 0 <= ty&&ty < m&&g[tx][ty] == '.'){//0<= tx, 不要写成0<tx
            totle++;
            dfs(tx, ty);
        }
    }
}
int main()
{
    while (true){
        cin >> m >> n;
        if (m + n == 0)break;
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                cin >> g[i][j];
                if (g[i][j] == '@'){
                    bx = i;
                    by = j;
                }
            }
        }
        totle = 1;
        dfs(bx, by);
        cout << totle << endl;
    }
    return 0;
}