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.
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)
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
4559613
分析:
只能够走‘.’,递归到最后,看一共有多少个可以走。
代码:
#include#include using namespace std;int m,n;int Next[4][2]= { {0,1},{1,0},{0,-1},{-1,0}};int bj[21][21];char a[21][21];int sum=0;void dfs(int x,int y){ int nx,ny; for(int i=0; i<4; i++) { nx=x+Next[i][0]; ny=y+Next[i][1]; if(a[nx][ny]=='.'&&nx>=0&&nx =0&&ny >n>>m) { if(m==0||n==0) break; for(i=0; i >a[i][j]; if(a[i][j]=='@') { x=i; y=j; } } memset(bj,0,sizeof(bj)); sum=1; dfs(x,y); cout< <