HDUOJ--------（1312）Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6905    Accepted Submission(s): 4384

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

 

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

 

Recommend

Eddy

 

 

 

 1 #include<cstdio>

 2 #include<cstdlib>

 3 #include<cstring>

 4 #include<deque>

 5 #include<iostream>

 6 #define maxn 25

 7 using namespace std;

 8 char maze[maxn][maxn];

 9 int w,h;

10 /*建立方向树*/

11 struct node

12 {

13   int x,y;

14 }start;

15 /*搜索方向*/

16 int dir[4][2]=

17 {

18     {0,1},

19     {0,-1},

20     {-1,0},

21     {1,0}

22 };

23 void bfs()

24 {

25     /*入队，出队用*/

26     deque<node>q;

27     /*暂存位置*/

28     node q1,q2;

29     q.push_back(start);

30     while(!q.empty())

31     {

32      q1=q.front ();

33      q.pop_front();

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

35      {

36          q2.x=q1.x+dir[i][0];

37          q2.y=q1.y+dir[i][1];

38          if(q2.x>h||q2.x<1||q2.y>w||q2.y<1||maze[q2.x][q2.y]=='#'||maze[q2.x][q2.y]=='p')   /*return ;*/

39               ;       /*啥也不干，就这么一个逗号....*/

40      else

41      {

42          if(maze[q2.x][q2.y]=='.')

43              maze[q2.x][q2.y]='p';   /*就暂时用P来代表占位置吧！*/

44         /*入队*/

45         q.push_back (q2);

46      }

47      }     

48     }

49 }

50 int main()

51 {

52     int i,j,ans;

53     while(scanf("%d%d",&w,&h),h+w)

54     {

55        getchar();

56        memset(maze,'\0',sizeof(maze));

57      for(i=1;i<=h;i++)

58      {

59        for(j=1;j<=w;j++)

60        {

61            scanf("%c",&maze[i][j]);

62            if(maze[i][j]=='@')

63            {

64                start.x=i;

65                start.y=j;

66                maze[i][j]='#';

67            }

68        }

69        getchar();

70      }

71        bfs();

72        ans=1;

73       for(i=1;i<=h;i++)

74       {

75           for(j=1;j<=w;j++)

76           {

77              

78               if(maze[i][j]=='p')

79                   ans++;

80           }

81       }

82       printf("%d\n",ans);

83     }

84   return 0;

85 }

View Code

bfs+匹配