Red and Black
Time Limit: 1000MS |
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Memory Limit: 30000K |
Total Submissions: 30914 |
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Accepted: 16846 |
Description
There is a rectangular and covered with square tiles. Each tile is colored either red or black. A man was standing on a black tile. From a tiles, he can move to one of the four adjacent tiles. But he can ' t move on red tiles, and 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 the positive integers W and H; W and H is the numbers of tiles in the X-and y-directions, respectively. W and H is not more than 20.
There is H more lines in the data set, and each of the 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 the 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.#..........#.#######. #.#.....#.. #.#.###.#.. #.#[email protected]#.#. #.#####.#.. #.......#.. #########............ 11 6..#. #.. #....#.. #.. #....#.. #.. ###.. #.. #.. #@...#.. #.. #....#.. #.. #.. 7 7..#.#....#.#. ###.###[email protected]###.###. #.#....#.#.. 0 0
Sample Output
4559613
Source
Japan 2004 Domestic resolution: Simple DFS.
#include <cstdio>int W, H;char s[25][25];int cnt;bool inroom (int x, int y) { return 0 <= x && x < h && 0 <= y && y < W;} void Dfs (int x, int y) { if (inroom (x, y) && s[x][y] = = '. ') { ++cnt; S[x][y] = ' # '; DFS (x-1, y); DFS (x+1, y); DFS (x, y-1); DFS (x, y+1);} } void Solve () { int x, y; for (int i = 0, i < H; ++i) for (int j = 0; J < W; ++j) if (s[i][j] = = ' @ ') { x = i; y = j; S[I][J] = '. '; Goto END; } END: cnt = 0; DFS (x, y); printf ("%d\n", CNT);} int main () {while (scanf ("%d%d", &w, &h), W) {for (int i = 0; i < h; ++i) scanf ("%s", S[i]); Solve (); } return 0;}
POJ 1979 Red and Black