POJ1979 Red and Black

POJ1979 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.

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#[email protected]#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
[email protected]
###.###
..#.#..
..#.#..
0 0

Sample Output

1
2
3
4
45
59
6
13

Answer

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40
#include <cstdio>

char maze[20][20];
int dx[4] = { -1, 0, 0, 1 };
int dy[4] = { 0, -1, 1, 0 };
int w, h, ans = 0;

void dfs(int i, int j) {
if (i < 0 || i >= h || j < 0 || j >= w)
return;
if (maze[i][j] == '#')
return;
++ans;
maze[i][j] = '#';
for (int x = 0; x < 4; ++x) {
dfs(i+dx[x], j+dy[x]);
}
}

int main() {
while (true) {
int sx, sy;
scanf("%d %d", &w, &h);
if (w == h && w == 0)
break;
for (int i = 0; i < h; ++i) {
scanf("%s", &maze[i]);
for(int j = 0; j < w; ++j) {
if (maze[i][j] == '@') {
sx = i;
sy = j;
}
}
}
dfs(sx, sy);
printf("%d\n", ans);
ans = 0;
}
return 0;
}

Analysis

  1. 二维矩阵中,有 '@','.','#',三种字符,其中 @ 是起点,# 是红色,. 是黑色,求@连通的最大最色格子数目。@ 本身是黑色。
  2. dfs 连通图。
# POJ

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