| Time Limit: 1000MS | Memory Limit: 32768KB | 64bit IO Format: %I64d & %I64u |
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)
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
Source
Asia 2004, Ehime (Japan), Japan Domestic
思路：裸DFS，扩展到一个点就将答案+1。
AC代码：
#include <cstdio>
using namespace std;
int w, h, ans;
char map[25][25];
const int dx[] = {0, 0, 1, -1},
dy[] = {1, -1, 0, 0};
struct node
{
int x, y;
}start;
bool input()
{
if (scanf("%d %d\n", &w, &h) == 2 && h)
{
for (int i = 0; i < h; i++)
scanf("%s\n", map[i]);
return 1;
}
return 0;
}
void dfs(node now)
{
if (now.x < 0 || now.x >= h || now.y < 0 || now.y >= w || map[now.x][now.y] == '#')
return;
map[now.x][now.y] = '#';
ans++;
node next;
for (int i = 0; i < 4; i++)
{
next.x = now.x + dx[i];
next.y = now.y + dy[i];
dfs(next);
}
}
void solve()
{
ans = 0;
for (int i = 0; i < h; i++)
for (int j = 0; j < w; j++)
if (map[i][j] == '@')
start = (node){i, j};
dfs(start);
printf("%d\n", ans);
}
int main()
{
while (input())
solve();
return 0;
}