hdu2952Counting Sheep
2015-06-06 20:44
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[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; bool mp[110][110]; char s[110]; bool visit[110][110]; int q[11000][2],l,r; int d1[] = {1,-1,0,0}; int d2[] = {0,0,1,-1}; void bfs(int a,int b) { l = r = 0; visit[a][b] = 1; q[r][0] = a,q[r++][1] = b; while(l<r) { int x = q[l][0],y = q[l++][1]; for(int j = 0;j<4;j++) { int xx = x+d1[j],yy = y+d2[j]; if(mp[xx][yy]&&!visit[xx][yy]) { visit[xx][yy] = 1; q[r][0] = xx,q[r++][1] = yy; } } } } int main() { int z,n,m,i,j,k; cin>>z; while(z--) { cin>>n>>m; memset(mp,0,sizeof(mp)); memset(visit,0,sizeof(visit)); for(i = 1;i<=n;i++) { scanf("%s",s); for(j = 0;j<m;j++) { mp[i][j+1] = (s[j] == '#')?1:0; } } int ans = 0; for(i = 1;i<=n;i++) { for(j = 1;j<=m;j++) { if(!visit[i][j]&&mp[i][j]) ans++,bfs(i,j); } } cout<<ans<<endl; } return 0; }
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