HDU：Counting Sheep

Counting Sheep

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 10

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

解题思路：

此题思路与油田问题一摸一样，但比油田问题更简单，上一题要搜索八个方位，而此题只需搜索四个方位。

只是方位数和符号所代表的含义变化了。其他代码均不变。

题目代码：

#include

#include

using namespace std;

int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};  //dir为direction的缩写

char mmap[103][103];

int m,n;

void _search(int x,int y)

{

        mmap[x][y]='.';

        for(int i=0;i<4;i++)

        {

            int dx=x+dir[i][0];

            int dy=y+dir[i][1];

            if(mmap[dx][dy]=='#'&&(dx>=1&&dx<=m)&&(dy>=1&&dy<=n))

            {

                _search(dx,dy);

            }

        }

}

int main()

{

    int sum,i,j,T;

    scanf("%d",&T);

    while(T--)                          

    {

        scanf("%d%d",&m,&n);             //m为行，n为列

        sum=0;

        for(i = 1;i <= m; i++)

            for(j = 1;j <= n; j++)

                scanf(" %c",&mmap[i][j]);

        for(i = 1;i <= m; i++)

            for(j = 1;j <= n; j++)

            {

                if(mmap[i][j]=='#')

                {

                    sum++;

                    _search(i,j);

                }

            }

        printf("%d\n",sum);

    }

    return 0;

}