ACM HDU-2952 Counting Sheep

 

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2060    Accepted Submission(s): 1359

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

 

题目来源：http://acm.hdu.edu.cn/showproblem.php?pid=2952

 

值得注意的是：bfs()函数里面定义的参数一定要定义成全局变量，如果用C语言的scanf来输入数据，要用getchar()来清除缓存区的'\n'。

 

#include<iostream>
using namespace std;

int arr[4][2]={0,1,1,0,0,-1,-1,0};    //分四个方向搜索 
char sheep[101][101];
int T,i,j,sum,h,w;

//搜索羊群 
void bfs(int a,int b)
{
    if(sheep[a][b]=='.')    return;            //遇到 '.'返回 
    if(a<0||b<0||a>=h||b>=w)    return;        //数组越界返回 
    sheep[a][b]='.';                        //记录，将每次找到的羊群变成'.'，下次循环直接跳过 
    for(int i=0;i<4;i++)
        bfs(a+arr[i][0],b+arr[i][1]);        //找到每只羊以后向四个方向搜索 
}

int main()
{    
    cin>>T;
    while(T--)
    {
        cin>>h>>w;
        for(i=0;i<h;i++)
            for(j=0;j<w;j++)
                 cin>>sheep[i][j];        //此处利用c++的输入方法，不用考虑缓冲区换行符的影响 
        sum=0;
        for(i=0;i<h;i++)
            for(j=0;j<w;j++)
            {
                if(sheep[i][j]=='#')
                {
                    ++sum;                //用sum来记录每次找到的羊群数量 
                    bfs(i,j);
                }
            }
        cout<<sum<<endl;
    }
    return 0;
}