Red and Black（DFS入门题）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4260    Accepted Submission(s): 2766
原题链接：点击打开链接

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

 

Recommend

Eddy

 

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#define LL long long
#define MAXI 2147483647
#define MAXL 9223372036854775807
#define eps (1e-8)
#define dg(i) cout << "*" << i << endl;
using namespace std;

int w, h, sx, sy, cnt;
char map[22][22];
char d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

void DFS(int i, int j)
{
    int ii, jj, k;
    for(k = 0; k < 4; k++)
    {
        ii = i + d[k][0];
        jj = j + d[k][1];
        if(ii > -1 && jj > -1 && ii < h && jj < w && map[ii][jj] == '.')
        {
            cnt++;
            map[ii][jj] = '#';
            DFS(ii, jj);
        }
    }
}

int main()
{
    int i, j;
    while(scanf("%d %d", &w, &h) && w)
    {
        cnt = 1;
        for(i = 0; i < h; i++)
        {
            scanf("%s", map[i]);
        }
        for(i = 0; i < h; i++)
        {
            for(j = 0; j < w; j++)
                if(map[i][j] == '@')
                {
                    sx = j;
                    sy = i;
                    i = h;
                    break;
                }
        }
        DFS(sy, sx);
        printf("%d\n", cnt);
    }
    return 0;
}