POJ 1979 Red and Black（DFS）

                         Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

题目大意：

        一个铺满黑砖和红砖的房间，一个刚开始站在黑砖上的人，在房间内移动，人只能站在黑砖上，并且每一次只能上下左右走一块，编写程序计算出人可以在这个房间走过多少黑砖。

'.' - a black tile （代表黑砖的位置）
'#' - a red tile （代表红砖的位置）
'@' - a man on a black tile(appears exactly once in a data set)（代表人的位置） 

代码如下：

#include <stdio.h>#include <string.h>//n,m:地图的长宽；二维数组（move[ ]） ：人每次移动的方向int n,m,num,move[4][2]={{1,0},{0,1},{-1,0},{0,-1}};//二维字符数组储存地图char map[25][25];//DFS深度搜索，找到人可以到达的黑砖数void DFS(int i,int j){    int k,x,y;    map[i][j]='#';    //循环四次，代表每一点的上下左右    for(k=0;k<4;k++)    {        x=i+move[k][0];        y=j+move[k][1];        if(x<n&&y<m&&x>=0&&y>=0&&map[x][y]=='.')        {             DFS(x,y);//沿一条路深搜下去             num++;        }    }}int main(){    int i,j,x,y;    while(scanf("%d%d",&m,&n))    {        getchar();//吃掉n后面的回车        if(m==0&&n==0)            return 0;        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                scanf("%c",&map[i][j]);                if(map[i][j]=='@')                {                    x=i;                    y=j;                }            }            getchar();//吃掉每行最后的回车        }        num=0;        DFS(x,y);        printf("%d\n",num+1);    }    return 0;}