链接：https://vjudge.net/problem/HDU-1312

Task

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:

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6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

output:

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2
3
4

45
59
6
13

Solution

建立两个二维数组，a用于存储输入字符串，b用于记录当前格子有没有被访问到。用递归实现深度优先搜索。 在DFS函数中，对上下左右四个方向讨论，对应4个if分支。每一层递归对当前节点的上下左右四个节点分别讨论，如果可以走（未到边界，不是#，未被走过）就走过去，如果不能走就退出一层递归，即退回到上一个节点，以此类推。 如下图是DFS搜索过程的可视化，左边是第4个输入示例，右边是更复杂的一个例子。 DFS

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#include<cstdio>
using namespace std;
// char array
char a[20][20];
// history, b[y][x]==0 for unseen cell
int  b[20][20];
int sum=0;
void DFS(int y, int x, int H, int W){
    sum++;
    printf("(%d,%d) -> %c\n",y,x,a[y][x]);
    b[y][x]=1;
    if(y>=1  && a[y-1][x]!='#' && b[y-1][x]==0)
        DFS(y-1,x,H,W);
    if(y<H-1 && a[y+1][x]!='#' && b[y+1][x]==0)
        DFS(y+1,x,H,W);
    if(x>=1  && a[y][x-1]!='#' && b[y][x-1]==0)
        DFS(y,x-1,H,W);
    if(x<W-1 && a[y][x+1]!='#' && b[y][x+1]==0)
        DFS(y,x+1,H,W);
}
int main(){
    int W,H,x_0,y_0;
    while(scanf("%d %d",&W,&H)!=EOF){
        if(W==0 && H==0)
            break;
        getchar();
        sum=0;
        for(int y=0; y<H; ++y){
            for(int x=0; x<W; ++x){
                scanf("%c",&(a[y][x]));
                b[y][x]=0;
                if(a[y][x]=='@'){
                    y_0=y;
                    x_0=x;
                }
            }
            getchar();
        }
        DFS(y_0,x_0,H,W);
        printf("%d\n",sum);
    }
    return 0;
}