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

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,'',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("%dn",ans);
83     }
84   return 0;
85 }

bfs+匹配

HDUOJ--------（1312）Red and Black

Red and Black