HDU 1312 Red and Black（bfs）

Red and Black

Time Limit:1000MS     Memory Limit:32768KB     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) 

 

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

题目简单翻译：

从‘@’点出发，问能到达的最多的点有多少，‘#’不可经过，计算结果包括‘@’。

 

解题思路：

广度优先搜索，直接求出到过多少个点。

 

代码：

 1 #include<cstdio>

 2 #include<cstring>

 3 

 4 using namespace std;

 5 int n,m,sx,sy;

 6 char mp[24][24];

 7 int vis[24][24];

 8 int dx[]={0,0,1,-1};

 9 int dy[]={1,-1,0,0};

10 bool check(int x,int y)

11 {

12     return x>=0&&x<n&&y>=0&&y<m;

13 }

14 struct node

15 {

16     int x,y;

17 }St[1000];

18 

19 int bfs()//手写的队列，队列的尾端就是到达的点的数量

20 {

21     memset(vis,0,sizeof vis);

22     int st=0,en=1;

23     vis[St[0].x][St[0].y]=1;

24     while(st<en)

25     {

26         node e=St[st++];

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

28         {

29             node w=e;

30             w.x=e.x+dx[i],w.y=e.y+dy[i];

31             if(check(w.x,w.y)&&vis[w.x][w.y]==0&&mp[w.x][w.y]!='#')

32             {

33                 vis[w.x][w.y]=1;

34                 St[en++]=w;

35             }

36         }

37     }

38     return en;

39 }

40 

41 int main()

42 {

43     while(scanf("%d%d",&m,&n)!=EOF&&(n||m))

44     {

45         for(int i=0;i<n;i++)

46         {

47             scanf("%s",mp[i]);

48             for(int j=0;j<m;j++)

49                 if(mp[i][j]=='@')

50                     St[0].x=i,St[0].y=j;

51         }

52         printf("%d\n",bfs());

53     }

54     return 0;

55 }

Red and Black