poj1979-Red and Black
Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 37141 | Accepted: 20181 |
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
dfs简单应用，用dfs深度搜索，每次遇到'.'就将cnt加一。我的dfs写了好久才对
#include
#include
#define maxn 21
using namespace std;
int vis[maxn][maxn];
char map[maxn][maxn];
int xx[4] = { 0,0,1,-1 };
int yy[4] = { 1,-1,0,0 };
int cnt;
int N, M;
int dfs(int x, int y) {
for (int i = 0; i < 4; i++) {
int nx = x + xx[i];
int ny = y + yy[i];
if (0 <= nx&&nx < N && 0 <= ny&&ny < M &&!vis[nx][ny]&& map[nx][ny] == '.') {
vis[nx][ny] = 1;
cnt++;
dfs(nx, ny);
}
}
return cnt;
}
int main() {
while (1) {
memset(vis, 0, sizeof(vis));
scanf("%d %d", &M, &N);
if (N == 0 || M == 0) break;
getchar();
int a, b;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
scanf("%c", &map[i][j]);
if (map[i][j] == '@') {
a = i; b = j;
}
}
getchar();
}
cnt = 0;
int ans = dfs(a, b);
printf("%d\n", ans+1);
}
return 0;
}