HDU 2952 Counting Sheep
題目連結:傳送門
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
題意:問#被.分為幾個部分。
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
char mp[101][101];
int a,b,s;
void dg(int i, int j)
{
if(mp[i][j]!='#'||i<0||j<0||i>=a||j>=b)
return;
else
{
mp[i][j]='*';
dg(i-1, j);
dg(i, j-1);
dg(i, j+1);
dg(i+1, j);
}
}
int main()
{
int i,j;
int w;
cin>>w;
while(w--)
{
cin>>a>>b;
s=0;
for(i=0;i<a;i++)
{
for(j=0;j<b;j++)
{
cin>>mp[i][j];
}
}
for(i=0;i<a;i++)
{
for(j=0;j<b;j++)
{
if(mp[i][j]=='#')
{
dg(i,j);
s++;
}
}
}
printf("%d\n",s);
}
return 0;
}