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

 

 

 

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<deque>
 #include<iostream>
 #define maxn 25
 using namespace std;
 char maze[maxn][maxn];
 int w,h;
 /*建立方向树*/
 struct node
 {
   int x,y;
 }start;
 /*搜索方向*/
 int dir[][]=
 {
     {,},
     {,-},
     {-,},
     {,}
 };
 void bfs()
 {
     /*入队，出队用*/
     deque<node>q;
     /*暂存位置*/
     node q1,q2;
     q.push_back(start);
     while(!q.empty())
     {
      q1=q.front ();
      q.pop_front();
      for(int i=;i<;i++)
      {
          q2.x=q1.x+dir[i][];
          q2.y=q1.y+dir[i][];
          if(q2.x>h||q2.x<||q2.y>w||q2.y<||maze[q2.x][q2.y]=='#'||maze[q2.x][q2.y]=='p')   /*return ;*/
               ;       /*啥也不干，就这么一个逗号....*/
      else
      {
          if(maze[q2.x][q2.y]=='.')
              maze[q2.x][q2.y]='p';   /*就暂时用P来代表占位置吧！*/
         /*入队*/
         q.push_back (q2);
      }
      }
     }
 }
 int main()
 {
     int i,j,ans;
     while(scanf("%d%d",&w,&h),h+w)
     {
        getchar();
        memset(maze,'\0',sizeof(maze));
      for(i=;i<=h;i++)
      {
        for(j=;j<=w;j++)
        {
            scanf("%c",&maze[i][j]);
            if(maze[i][j]=='@')
            {
                start.x=i;
                start.y=j;
                maze[i][j]='#';
            }
        }
        getchar();
      }
        bfs();
        ans=;
       for(i=;i<=h;i++)
       {
           for(j=;j<=w;j++)
           {

               if(maze[i][j]=='p')
                   ans++;
           }
       }
       printf("%d\n",ans);
     }
   return ;
 }

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