[POJ] 1979 Red and Black

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 21102		Accepted: 11267

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

Source

Japan 2004 Domestic

 

题解：广度优先搜索，从@处出发统计能遍历的black tile的个数，@也算一个。

 

代码：

 1 #include<stdio.h>

 2 #include<stdbool.h>

 3 #include<string.h>

 4 int i,j,n,m,sum,head,tail,sum,x,y,

 5     q[100000][2],

 6     dx[4]={0,0,1,-1},

 7     dy[4]={1,-1,0,0};

 8 

 9 char a[100][100];

10 

11 bool can[100][100];

12 

13 int 

14 pre()

15 {

16     memset(q,0,sizeof(q));

17     memset(can,false,sizeof(can));

18     sum=1;

19     return 0;

20 }

21 

22 int

23 init()

24 {

25     for(i=1;i<=n;i++)

26     {

27         scanf("%s",&a[i]);

28         for(j=0;j<m;j++)

29         {

30             if(a[i][j]=='.')

31             can[i][j]=true;

32             

33             if(a[i][j]=='@')

34             {

35                 x=i;y=j;

36                 can[i][j]=true;

37             }

38         }

39     }

40     

41     return 0;

42 }

43 

44 int 

45 bfs()

46 {

47     int i,xi,yi,xx,yy;

48     

49     head=0;tail=1;

50     can[x][y]=false;

51     q[1][0]=x;q[1][1]=y;

52     

53     while(head!=tail)

54     {

55         head=head%100000+1;

56         xi=q[head][0];

57         yi=q[head][1];

58         

59         for(i=0;i<4;i++)

60         {

61             xx=xi+dx[i];

62             yy=yi+dy[i];

63             if(can[xx][yy])

64             {

65                 tail=tail%100000+1;

66                 q[tail][0]=xx;

67                 q[tail][1]=yy;

68                 can[xx][yy]=false;

69                 sum++;

70             }

71         }

72     }

73     printf("%d\n",sum);

74     return 0;

75     

76     

77 }

78              

79 int 

80 main()

81 {

82     while(true)

83     {

84         pre();

85         scanf("%d%d\n",&m,&n);

86         if(m==0&&n==0) break;

87         init();

88         bfs();

89     }

90     

91    

92     return 0;

93 }