HDUOJ----2952Counting Sheep

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1782    Accepted Submission(s): 1170

Problem Description

A while ago I had trouble sleeping. I used to lie awake, staring at the ceiling, for hours and hours. Then one day my grandmother suggested I tried counting sheep after I'd gone to bed. As always when my grandmother suggests things, I decided to try it out. The only problem was, there were no sheep around to be counted when I went to bed.


Creative as I am, that wasn't going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while . is grass (or whatever you like, just not sheep). To make the counting a little more interesting, I also decided I wanted to count flocks of sheep instead of single sheep. Two sheep are in the same flock if they share a common side (up, down, right or left). Also, if sheep A is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps A and C are in the same flock.


Now, I've got a new problem. Though counting these sheep actually helps me fall asleep, I find that it is extremely boring. To solve this, I've decided I need another computer program that does the counting for me. Then I'll be able to just start both these programs before I go to bed, and I'll sleep tight until the morning without any disturbances. I need you to write this program for me.

 

 

Input

The first line of input contains a single number T, the number of test cases to follow.

Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. Then follows H lines, each containing W characters (either # or .), describing that part of the grid.

 

 

Output

For each test case, output a line containing a single number, the amount of sheep flock son that grid according to the rules stated in the problem description.

Notes and Constraints
0 < T <= 100
0 < H,W <= 100

 

 

Sample Input

2 4 4 #.#. .#.# #.## .#.# 3 5 ###.# ..#.. #.###

 

 

Sample Output

6 3

 

 

Source

IDI Open 2009

简单的搜索...

代码：

 1 //简单的搜索

 2 #include<cstdio>

 3 #include<queue>

 4 #include<iostream>

 5 using namespace std;

 6 const int maxn=101;

 7 char map[maxn][maxn];

 8 typedef struct

 9 {

10     int x,y;

11 }po;

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

13 int main()

14 {

15     int n,m,t,i,j,k,ans;

16     queue<po>tem;

17     scanf("%d",&t);

18     while(t--)

19     {

20         ans=0;

21       scanf("%d%d",&n,&m);

22       for(i=0;i<n;i++)

23           scanf("%s",map[i]);

24       for(i=0;i<n;i++)

25       {

26           for(j=0;j<m;j++)

27           {

28               if(map[i][j]=='#')

29               {

30                   ans++;

31                   map[i][j]='.';

32                   po st={i,j};

33                   tem.push( st );

34                   while(!tem.empty())

35                   {

36                       po en=tem.front();

37                       tem.pop();

38                       for(k=0;k<4;k++)

39                       {

40                           if(map[en.x+dir[k][0]][en.y+dir[k][1]]=='#')

41                           {

42                               map[en.x+dir[k][0]][en.y+dir[k][1]]='.';

43                               po sa={en.x+dir[k][0],en.y+dir[k][1]};

44                               tem.push(sa);

45                           }

46                       }

47                   }

48               }

49           }

50       }

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

52     }

53  return 0;

54 }

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