HDU1312 Red and Black
Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9732 Accepted Submission(s): 6060
Problem 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.
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)
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 <stdio.h> #include <string.h> #define maxn 22 int n, m, ans; const int mov[][2] = {1, 0, -1, 0, 0, 1, 0, -1}; char map[maxn][maxn]; bool check(int x, int y){ return x >= 0 && y >= 0 && x < n && y < m && map[x][y] != '#'; } void DFS(int x, int y) { int a, b, i; for(i = 0; i < 4; ++i){ a = x + mov[i][0]; b = y + mov[i][1]; if(check(a, b)){ ++ans; map[a][b] = '#'; DFS(a, b); } } } int main() { int i, j, x, y; while(scanf("%d%d", &m, &n) == 2 && (m || n)){ for(i = 0; i < n; ++i){ getchar(); for(j = 0; j < m; ++j){ map[i][j] = getchar(); if(map[i][j] == '@'){ x = i; y = j; } } } map[x][y] = '#'; ans = 1; DFS(x, y); printf("%d\n", ans); } return 0; }