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POJ 1979 Red and Black（简单DFS）
Red and Black
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
4559613
最简单常规的深度优先搜索（DFS）题，与POJ2386这个题很类似。不再详细解释，直接上代码：
#include<iostream>#include<algorithm>
#include<cstring>using namespace std;char maze[25][25];bool flag[25][25];int n,m,dir[4][2] = { {1,0},{-1,0},{0,1},{0,-1} },ans;void dfs(int x, int y){ if (x<=0||x > n ||y<=0|| y > m|| maze[x][y] == ‘#‘) return; if (flag[x][y])//这个位置未被标记，之前没被走过 { ans++; flag[x][y] = false; } maze[x][y] = ‘#‘;//对当前位置置换成#，避免下次对这个位置重复DFS for (int i = 0; i < 4; i++)//向四个方向DFS { int nx = x + dir[i][0], ny = y + dir[i][1]; dfs(nx, ny); } return ;}int main(){ while (cin >> m >> n,(m||n)) { memset(flag, true, sizeof(flag)); int sx, sy; for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) { cin >> maze[i][j]; if (maze[i][j] == ‘@‘) { sx = i; sy = j; } } ans = 0; dfs(sx, sy); cout << ans << endl; } return 0;}
POJ 1979 Red and Black（简单DFS）
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