HDU 1312 Red and Black (搜索)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9795    Accepted Submission(s): 6103

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

 

 

搜索求出这个人能走多少黑色格子

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<cstring>

 4 #include<stdlib.h>

 5 #include<algorithm>

 6 using namespace std;

 7 const int MAXN=100;

 8 int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

 9 char map[MAXN][MAXN],s[MAXN];

10 int vis[MAXN][MAXN];

11 int ans,n,m;

12 void DFS(int x,int y)

13 {

14     vis[x][y]=1;

15     ans++;

16     for(int i=0;i<4;i++)

17     {

18         int xx=x+dir[i][0];

19         int yy=y+dir[i][1];

20         if(0<=xx&&xx<n&&0<=yy&&yy<m&&!vis[xx][yy]&&map[xx][yy]!='#')

21             DFS(xx,yy);

22     }

23 }

24 int main()

25 {

26     //freopen("in.txt","r",stdin);

27     int x,y;

28     while(scanf("%d %d%*c",&m,&n)&&(n||m))

29     {

30         for(int i=0;i<n;i++)

31         {

32             for(int j=0;j<m;j++)

33             {

34                 scanf("%c",&map[i][j]);

35                 if(map[i][j]=='@')

36                 {

37                     x=i;

38                     y=j;

39                 }

40             }

41             getchar();

42         }

43         memset(vis,0,sizeof(vis));

44         ans=0;

45         DFS(x,y);

46         printf("%d\n",ans);

47     }

48     return 0;

49 }

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