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 }
View Code
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