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;
}

 

分类: ACM/ICPC

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