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);}}