DFS解決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
View Code
#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個方向的走法下，上，左， 右 int n, m; int count;//記錄走過的總數 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; }
posted on 2011-09-23 22:38 More study needed. 閱讀(150) 評論(0) 收藏 舉報
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