杭电1312--Red and Black（Dfs）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13090    Accepted Submission(s): 8116

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

 

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

 

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//比较基础的搜索题， 但是做了很久。→ → ！！

#include <cstdio>
#include <iostream>
#include <cstring>
#define M 30
using namespace std;

int ans;
int x, y;
char map[M][M];
int dx[4]={0, 0, 1, -1}; 
int dy[4]={1, -1, 0, 0};
int a, b;
void dfs(int x, int y)
{
    int m, n;
    ans++;
    map[x][y] = '@';
    //printf("%d %d


", x, y);
    for(int i = 0; i < 4; i++) //  
    {
        m = x + dx[i];
        n = y + dy[i];
        //printf("%d %d
", m, n); 
        if(m >= 0 && m < b && n >= 0 && n < a && map[m][n] == '.')   //限制边界。 
        {
        //    printf("1");
        //    printf(" %d %d
", m, n);
            dfs(m, n);
        }
    }
}

int main()
{
    int i, j;
    while(~scanf("%d %d", &a, &b), a|b)    
    {
        ans = 0;
        for(i=0; i<b; i++)
        {
            for(j=0; j<a; j++)
            {
                cin >> map[i][j];
                if(map[i][j] == '@')
                {
                    x = i;
                    y = j;
                }
            }
        }
        //printf("%d %d

", x, y);
        dfs(x, y);
        printf("%d
", ans);
    }
    return 0;
} 

注意：

void dfs(int x, int y)
{
    int nx, ny;
    ant++;
    map[x][y] = '@';
    for(int i = 0; i < 4; i++)
    {
        nx = x + acx[i];
        ny = y + acy[i];
        if(nx >= 0 && nx < m && ny >= 0 && ny < n && map[nx][ny] == '.')
        {
            dfs(nx, ny);    
        }  
    }
}