HDU5952-Counting Sheep
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Counting Sheep
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3061 Accepted Submission(s): 2068
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.
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.
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
Notes and Constraints
0 < T <= 100
0 < H,W <= 100
Sample Input
24 4#.#..#.##.##.#.#3 5###.#..#..#.###
Sample Output
63
Source
IDI Open 2009
#include <iostream>#include <string.h>#include <stdio.h>using namespace std;#define N 105int m,n;int dir[4][2]= {{0,1},{1,0},{0,-1},{-1,0}};int visited[N][N];char map[N][N];int sum=0;void DFS(int x,int y){ int nx,ny; for(int i=0; i<4; i++) { nx=x+dir[i][0]; ny=y+dir[i][1]; if(map[nx][ny]=='#'&&nx>=0&&nx<m&&ny>=0&&ny<n&&visited[nx][ny]==0) { visited[nx][ny]=1; DFS(nx,ny); } }}int main(){ int t; scanf("%d",&t); while(t--) { cin>>m>>n; if(m==0&&n==0) break; for(int i=0; i<m; i++) { for(int j=0; j<n; j++) cin>>map[i][j]; } memset(visited,0,sizeof(visited)); sum=0; for(int i=0; i<m; i++) { for(int j=0; j<n; j++) { if(map[i][j]=='#'&&visited[i][j]==0) { visited[i][j]=1; DFS(i,j); sum++; } } } cout<<sum<<endl; } return 0;}
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