HDOJ 2952 Counting Sheep
Counting Sheep
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 2231 Accepted Submission (s): 1474
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
Zero <T <= 100
0 <H, W <= 100
Sample Input
24 4#.#..#.##.##.#.#3 5###.#..#..#.###
Sample Output
63
Source IDI open2009
Question: Give you a picture, and calculate that there are fewer.
Problem: recursive marking is good... DFS does not require backtracking.
AC code:
#include
#include
#define N 105using namespace std;string str[N];int t,m,n;int dir[][2]={ {0,1},{0,-1},{1,0},{-1,0}};void dfs(int x,int y ){ for(int i=0;i<4;i++){ int dx=x+dir[i][0],dy=y+dir[i][1]; if(dx>=0&&dx
=0&&dy
>t; while(t--){ int sum=0; cin>>m>>n; for(int i=0;i
>str[i]; for(int i=0;i
