hdu 1312 Red and Black

http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6095    Accepted Submission(s): 3883


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
 

 dfs模板题 

 1 #include<stdio.h>
 2 #include<string.h>
 3 char map[25][25];
 4 int dir[4][2]={1,0,-1,0,0,1,0,-1};//方向数组
 5 int vis[25][25];//标记数组
 6 int n,m,count;//行，列，计数器
 7 void dfs(int x,int y)//标准的DFS模板
 8 {
 9      int i;
10      for(i=0;i<4;i++)
11      {
12           int xx=x+dir[i][0];
13           int yy=y+dir[i][1];
14           if(xx>=0&&xx<n&&yy>=0&&yy<m&&!vis[xx][yy]&&map[xx][yy]!='#')
15           {
16                count++;
17                vis[xx][yy]=1;
18                dfs(xx,yy);
19           }
20      }
21    //  return ;
22 }
23 int main()
24 {
25     int i,j;
26     while(scanf("%d %d",&m,&n)!=EOF)
27     {
28          if(m==0||n==0)
29          break;
30          memset(map,0,sizeof(map));//因为是多组输入所以每次得清0
31          memset(vis,0,sizeof(vis));
32          count=1;
33          for(i=0;i<n;i++)
34          scanf("%s",map[i]);
35          for(i=0;i<n;i++)
36             for(j=0;j<m;j++)
37             {
38                  if(map[i][j]=='@')
39                  {
40                       map[i][j]='#';
41                       vis[i][j]=1;
42                       dfs(i,j);
43                  }
44             }
45             printf("%d\n",count);
46     }
47      return 0;
48 }
View Code

 

posted @ 2013-07-17 19:54  SprayT  阅读(162)  评论(0)  编辑  收藏  举报