《挑战》2.1 POJ POJ 1979 Red and Black （简单的DFS）

B - Red and Black

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 1979

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

45

59

6

13

题目大意：

　　就是说，对于一个h*w的网格，每次从‘@’开始，求最多能走多少个‘.’，每次只能够向四个方向进行扩展。

解题思路：

　　直接dfs，然后用book数组随时记录目前这个点有没有走过，如果这个点是'.'并且没有走过的话，那么我们就res++，由于是多组数据，每次输出结束后，记着memset,

复杂度：

　　O(4*w*h)

代码： 

 1 # include<cstdio>

 2 # include<iostream>

 3 # include<cstring>

 4 

 5 using namespace std;

 6 

 7 # define MAX 23

 8 

 9 char grid[MAX][MAX];

10 int book[MAX][MAX];

11 int nxt[4][2] = {{1,0},{0,-1},{-1,0},{0,1} };

12 int res;

13 int w,h;

14 

15 int can_move ( int x,int y )

16 {

17     if ( x>=0&&x<h&&y>=0&&y<w )

18     {

19         if ( grid[x][y]=='.'&&book[x][y]==0 )

20             return 1;

21     }

22     return 0;

23 }

24 

25 void dfs ( int x,int y )

26 {

27     book[x][y] = 1;

28     for ( int i = 0;i < 4;i++ )

29     {

30         int n_x = x+nxt[i][0], n_y = y+nxt[i][1];

31         if ( can_move ( n_x,n_y ) )

32         {

33             res++;

34             book[n_x][n_y] = 1;

35             dfs(n_x,n_y);

36         }

37     }

38 }

39 

40 int main(void)

41 {

42     while ( scanf("%d%d",&w,&h)!=EOF )

43     {

44         res = 1;

45         if ( w==0&&h==0 )

46             break;

47         for ( int i = 0;i < h;i++ )

48             scanf("%s",grid[i]);

49         for ( int i = 0;i < h;i++ )

50         {

51             for ( int j = 0;j < w;j++ )

52             {

53                 if ( grid[i][j]=='@' )

54                 {

55                     book[i][j] = 1;

56                     dfs(i,j);

57                 }

58             }

59         }

60 

61         printf("%d\n",res);

62         memset(book,0,sizeof(book));

63     }

64 

65 

66 

67     return 0;

68 }