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：

 #include<cstdio>
 #include<cstring>

 using namespace std;
 int n,m,sx,sy;
 char mp[][];
 int vis[][];
 int dx[]={,,,-};
 int dy[]={,-,,};
 bool check(int x,int y)
 {
     return x>=&&x<n&&y>=&&y<m;
 }
 struct node
 {
     int x,y;
 }St[];

 int bfs()//手写的队列，队列的尾端就是到达的点的数量
 {
     memset(vis,,sizeof vis);
     int st=,en=;
     vis[St[].x][St[].y]=;
     while(st<en)
     {
         node e=St[st++];
         for(int i=;i<;i++)
         {
             node w=e;
             w.x=e.x+dx[i],w.y=e.y+dy[i];
             if(check(w.x,w.y)&&vis[w.x][w.y]==&&mp[w.x][w.y]!='#')
             {
                 vis[w.x][w.y]=;
                 St[en++]=w;
             }
         }
     }
     return en;
 }

 int main()
 {
     while(scanf("%d%d",&m,&n)!=EOF&&(n||m))
     {
         for(int i=;i<n;i++)
         {
             scanf("%s",mp[i]);
             for(int j=;j<m;j++)
                 if(mp[i][j]=='@')
                     St[].x=i,St[].y=j;
         }
         printf("%d\n",bfs());
     }
     return ;
 }

Red and Black