poj 1979 &&hdu 1312Red and Black (bfs)

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

6 9

....#.

.....#

......

......

......

......

......

#@...#

.#..#.

11 9

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

...@...

###.###

..#.#..

..#.#..

0 0

Sample Output

45

59

6

13

solution：

bfs水题

#include<cstdio>
#include<queue>
#include<iostream>
using namespace std;
int n, m,vis[50][50];
char map[50][50];
int dx[5] = { -1, 0, 0, 1 }, dy[5] = { 0, -1, 1, 0 };
struct node{
int x, y;
}now,no;
int main()
{
while (scanf("%d%d", &m, &n) && n + m)
{
queue<node>q;
getchar();
for (int i = 0; i < 50; i++)
for (int j = 0; j < 50; j++)
vis[i][j] = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
scanf("%c", &map[i][j]);
if (map[i][j] == '@'){ vis[i][j] = 1; no.x = i; no.y = j; q.push(no); }
}
getchar();
}
int ans = 1;
while (!q.empty())
{
now = q.front(); q.pop();
for (int i = 0; i < 4; i++)
{
no.x = now.x + dx[i]; no.y = now.y + dy[i];
if (no.x < 0 || no.x >= n || no.y < 0 || no.y >= m)continue;
if (map[no.x][no.y] == '.'&&vis[no.x][no.y] == 0)
{
vis[no.x][no.y] = 1; q.push(no); ans++;
}
}
}
printf("%d\n", ans);
}
}