HDU 1312 Red and Black

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

45
59
6
13
题意：输入两个数N和M代表有一个N*M的矩阵，然后有M行，每行有N个字符，其中‘#’表示不能通过，‘.’表示可以通过，‘@’表示初始位置，求从初始位置开始最多可以经历多少个点。
题解：
这是一道典型的深搜，直接递归解决问题，或者也可以用栈。

#include<stdio.h>#include<string.h>#include<stdlib.h>int a[21][21];//用来表是点是否经历过char Map[22][22];//N*M的矩阵，储存输入的字符//上面两个也可以只使用一个，只是在经过一个点时把该点的字符改为‘#’int res;//表示经历了多少个点int s[4][2]={{1,0},{-1,0},{0,1},{0,-1}};//表示四个可以走的方向int N,M;void dfs(int x,int y){   a[y][x]=1;   res++;//每经过一个点就把次数加一   for(int i=0;i<4;i++)//遍历四个方向   {       int nx=x+s[i][0];       int ny=y+s[i][1];       if(nx>=1&&nx<=N&&ny>=1&&ny<=M&&Map[ny][nx]=='.'&&a[ny][nx]==0)//判断当前点是否可以通过和是否在合法范围        {            dfs(nx,ny);        }   }}int main(){    int q,w;    while(scanf("%d %d",&N,&M)!=EOF)    {        res=0;//将初始经历点数置为初始值0        memset(a,0,sizeof(a));        if(N==0&&M==0)            break;        for(int i=1;i<=M;i++)        {            getchar();            for(int j=1;j<=N;j++)        {            scanf("%c",&Map[i][j]);            if(Map[i][j]=='@')            {                q=i;                w=j;            }//记录下初始位置        }        }        dfs(w,q);//从初始位置开始搜索        printf("%d\n",res);    }    return 0;}

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