Poj 1979 red and black

Red and black

Time limit:1000 ms		Memory limit:30000 K
Total submissions:22300		Accepted:12041

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 ctions, respectively. W and H are not more than 20.

There are h more lines in the data set, each of which between des 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 shocould 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

Question:
First, this is an image. '.' indicates the black brick, '#' indicates the red brick, and '@' indicates the black brick where the hero stays. Red bricks cannot be taken, but they can only follow the Black bricks. Each time you take a step, ask the maximum number of Black bricks that can be taken!

Golden DFS;
Pay attention to the boundary and recursive end conditions!

Code:

# Include <iostream> # include <cstdio> # include <cstring> # include <queue> using namespace STD; # define maxn 25 # define mm 10000000 char s [maxn] [maxn]; int W [maxn] [maxn]; // mark the passed vertex int Sx, Sy; int conut; // count int n, m; int XX [] = {0, 1,-1}; int YY [] = {1,-1, 0 }; int DFS (int x, int y) {If (s [x] [Y] = '#' | W [x] [Y]) return 0; conut ++; W [x] [Y] = 1; for (INT I = 0; I <4; I ++) {int xi = x + XX [I]; int YI = Y + YY [I]; If (xi> = 0 & xi <n & Yi> = 0 & Yi <m) DFS (XI, Yi) ;}return conut ;}int main () {While (~ Scanf ("% d", & M, & N) {If (M = 0 & n = 0) break; memset (W, 0, sizeof (w); conut = 0; For (INT I = 0; I <n; I ++) {scanf ("% s", s [I]); // input for (Int J = 0; j <m; j ++) if (s [I] [J] = '@') // mark the start point {SX = I; Sy = J ;}} int K = DFS (sx, Sy); printf ("% d \ n", k );} return 0 ;}