DFS解决POJ 1979

淡淡的烟草味﹌ 2021-11-11 11:50 225阅读 0赞

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

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

#include<iostream>
    using namespace std;
    int used[25][25];
    char map[25][25];
    int d[4][2] = {
          {-1,0}, {
          1,0}, {
          0,-1}, {
          0,1}};//4个方向的走法下，上，左， 右
    int n, m;
    int count;//记录走过的总数
    void dfs(int i, int j)
    {
        count++;
        used[i][j]=1;
        int ii;
        for(ii=0; ii<4; ii++)
        {
            if(i+d[ii][0]>=0 && i+d[ii][0]<n && j+d[ii][1]>=0 && j+d[ii][1]<m && map[i+d[ii][0]][j+d[ii][1]]=='.' && !used[i+d[ii][0]][j+d[ii][1]])
            {
                dfs(i+d[ii][0],j+d[ii][1]);
            }
        }
    }
    int main()
    {
        int start_i, start_j, i, j;
        while(cin>>m>>n && (n+m))
        {
            memset(used, 0, sizeof(used));
            for(i=0; i<n; i++)
            {
                for(j=0; j<m; j++)
                {
                    cin>>map[i][j];
                    if(map[i][j]=='@')
                    {
                        start_i = i;
                        start_j = j;
                    }
                }
            }
            count = 0;
            dfs(start_i, start_j);
            cout<<count<<endl;
        }
        return 0;
    }

转载于:https://www.cnblogs.com/o8le/archive/2011/09/23/2186746.html

[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
[ExpandedBlockStart.gif]: /images/20211109/99198a224ad343dea5509e8ec10d3bd8.png