POJ 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 30914		Accepted: 16846

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

 

 

 

解析：简单DFS。

 

 

 

#include <cstdio>

int w, h;
char s[25][25];
int cnt;

bool inRoom(int x, int y)
{
    return 0 <= x && x < h && 0 <= y && y < w;
}

void dfs(int x, int y)
{
    if(inRoom(x, y) && s[x][y] == '.'){
        ++cnt;
        s[x][y] = '#';
        dfs(x-1, y);
        dfs(x+1, y);
        dfs(x, y-1);
        dfs(x, y+1);
    }
}

void solve()
{
    int x, y;
    for(int i = 0; i < h; ++i)
        for(int j = 0; j < w; ++j)
            if(s[i][j] == '@'){
                x = i;
                y = j;
                s[i][j] = '.';
                goto END;
            }
    END:
    cnt = 0;
    dfs(x, y);
    printf("%d
", cnt);
}

int main()
{
    while(scanf("%d%d", &w, &h), w){
        for(int i = 0; i < h; ++i)
            scanf("%s", s[i]);
        solve();
    }
    return 0;
}