Red and Black
Time Limit: 2 Seconds Memory Limit: 65536 KB
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. Wand 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
#include <iostream>
#include <queue>
using namespace std;
struct node {
int r;
int l;
};
int m, n, sr, sl;
int dir[2][4] = {{-1, 1, 0, 0}, {0, 0, -1, 1}};
int visited[20][20];
int BFS();
int main()
{
int i, j;
char str[21];
while(cin >> m >> n) {
if(m == 0 || n == 0) {
break;
}
for(i = 0; i < n; i++) {
cin >> str;
for(j = 0; j < m; j++) {
if(str[j] == '@') {
sr = i;
sl = j;
visited[i][j] = 1;
}
else if(str[j] == '.') {
visited[i][j] = 0;
} else {
visited[i][j] = 1;
}
}
}
int count = BFS();
cout << count << endl;
}
return 0;
}
int BFS()
{
int i, count;
queue<node>Q;
node N, temp;
N.r = sr;
N.l = sl;
count = 1;
Q.push(N);
while(!Q.empty()) {
temp = Q.front();
Q.pop();
for(i = 0; i < 4; i++) {
N = temp;
N.r += dir[0][i];
N.l += dir[1][i];
if(N.r < 0 || N.r >= n ||
N.l < 0 || N.l >= m) {
continue;
} else {
if(!visited[N.r][N.l]) {
count++;
visited[N.r][N.l] = 1;
Q.push(N);
}
}
}
}
return count;
}