搜索入门-----HDU1312

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

大致意思是：    @是这个人的坐标，可以走 ' . ' ,   但不能走’ # ‘, 问一共多少走法

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char s[25][25];int sum, a, b;void dfs(int m, int n){if(s[m][n] == '#')//遇见 # 返回 return ;if(m >= b || m < 0 || n >= a || n < 0)//超出范围返回 return ;sum++;s[m][n] = '#';//走过的不能再走 dfs(m + 1, n);dfs(m, n + 1);dfs(m - 1, n);dfs(m, n - 1);}int main(){int m, n;while(scanf("%d%d", &a, &b) == 2 && a && b){int ok = 0;for(int i = 0; i < b; i++){scanf("%s", s[i]);if(!ok){for(int j = 0; j < a; j++){if(s[i][j] == '@'){m = i;n = j;ok = 1;}}}}sum = 0;dfs(m, n);printf("%d\n", sum);}return 0;}