HDU 1312 Red and Black

HDU 1312 Red and Black



Problem 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

这道题的意思是说，一个房间里有两张瓷砖，一种红色（用#表示），一种黑色（用 . 表示）。一个人在房间里走，只能在黑色瓷砖上走，一次只能达到相邻的四块瓷砖，问，这个人做多能走过多少块瓷砖？

使用深搜，一步一步走就行了，水 ￣へ￣

直接上代码了 (～￣▽￣)～

[code]
 #include <stdio.h>
 #include <iostream>
 #include <string.h>
using namespace std;
int w ,h, ans;
char a[25][25];
int vis[25][25];
int dir[4][2]={{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
void dfs(int x, int y)
{
    int nx, ny;
    for (int i = 0; i < 4; i++)
    {
        nx = x + dir[i][0];
        ny = y + dir[i][1];
        if (a[nx][ny] == '.' && nx >= 0 && nx < h && ny >= 0 && ny < w && !vis[nx][ny])
        {
            vis[nx][ny] = 1;
            dfs(nx, ny);
        }
    }
}
int main()
{
 #ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
 #endif
    int i, j;
    while(scanf("%d%d", &w, &h), w)
    {
        memset(a, 0, sizeof(a));
        for(i = 0; i < h; i++)
        {
            scanf("%s", a[i]);    //gets{a[i])为什么WA???    why???
        }
        ans = 0;
        memset(vis, 0, sizeof(vis));
        for (i = 0; i < h; i++)
        {
            for (j = 0; j < w; j++)
            {
                if (a[i][j] == '@')
                {
                    vis[i][j] = 1;
                    dfs(i, j);
                    break;
                }
            }
        }
        for (i = 0; i < h; i++)
        {
            for (j = 0; j < w; j++)
            {
                ans += vis[i][j];
            }
        }
        cout << ans << endl;
    }
    return 0;
}