DFS解决POJ 1979
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.
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.
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
#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;
}