[POJ] 1979 Red and Black

Red and Black
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 21102   Accepted: 11267

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) 
The end of the input is indicated by a line consisting of two zeros. 

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

Japan 2004 Domestic
 
题解：广度优先搜索，从@处出发统计能遍历的black tile的个数，@也算一个。
 
代码：
 1 #include<stdio.h>
 2 #include<stdbool.h>
 3 #include<string.h>
 4 int i,j,n,m,sum,head,tail,sum,x,y,
 5     q[100000][2],
 6     dx[4]={0,0,1,-1},
 7     dy[4]={1,-1,0,0};
 8 
 9 char a[100][100];
10 
11 bool can[100][100];
12 
13 int 
14 pre()
15 {
16     memset(q,0,sizeof(q));
17     memset(can,false,sizeof(can));
18     sum=1;
19     return 0;
20 }
21 
22 int
23 init()
24 {
25     for(i=1;i<=n;i++)
26     {
27         scanf("%s",&a[i]);
28         for(j=0;j<m;j++)
29         {
30             if(a[i][j]=='.')
31             can[i][j]=true;
32             
33             if(a[i][j]=='@')
34             {
35                 x=i;y=j;
36                 can[i][j]=true;
37             }
38         }
39     }
40     
41     return 0;
42 }
43 
44 int 
45 bfs()
46 {
47     int i,xi,yi,xx,yy;
48     
49     head=0;tail=1;
50     can[x][y]=false;
51     q[1][0]=x;q[1][1]=y;
52     
53     while(head!=tail)
54     {
55         head=head%100000+1;
56         xi=q[head][0];
57         yi=q[head][1];
58         
59         for(i=0;i<4;i++)
60         {
61             xx=xi+dx[i];
62             yy=yi+dy[i];
63             if(can[xx][yy])
64             {
65                 tail=tail%100000+1;
66                 q[tail][0]=xx;
67                 q[tail][1]=yy;
68                 can[xx][yy]=false;
69                 sum++;
70             }
71         }
72     }
73     printf("%d\n",sum);
74     return 0;
75     
76     
77 }
78              
79 int 
80 main()
81 {
82     while(true)
83     {
84         pre();
85         scanf("%d%d\n",&m,&n);
86         if(m==0&&n==0) break;
87         init();
88         bfs();
89     }
90     
91    
92     return 0;
93 }

 

posted @ 2014-04-03 22:05  SXISZERO  阅读(225)  评论(0)  编辑  收藏  举报