HDU 2952 Counting Sheep

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1727    Accepted Submission(s): 1121

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

 

Source

IDI Open 2009

 

Recommend

gaojie

 

思路：

 

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <cstring>
using namespace std;
int hash[110][110];
char map[110][110];
int bfs[4][2] = {1,0,-1,0,0,1,0,-1};
int t,h,w;
int sum;
int a,b;
struct Node
{
    int x,y;
};
void BFS()
{
    queue < Node > q;
    Node top;
    top.x = a;top.y = b;
    q.push(top);
    while(!q.empty())
    {
        Node temp;
        temp = q.front();
        q.pop();
        for(int i = 0;i < 4;i ++)
        {
            int x = temp.x + bfs[i][0],y = temp.y + bfs[i][1];
            if(map[x][y] == '#' && hash[x][y] == 0 &&
            x >= 1 && x <= h && y >= 1 && y <= w)
            {
                Node xin;
                xin.x = x;xin.y = y;
                hash[x][y] = 1;
                q.push(xin);
            }
        }
    }
}
int main()
{
    scanf("%d",&t);
    while(t --)
    {
        memset(hash,0,sizeof(hash));
        scanf("%d%d",&h,&w);
        for(int i = 1;i <= h;i ++)
            for(int j  = 1;j <= w;j ++)
                scanf(" %c",&map[i][j]);
        sum = 0;
        for(int i = 1;i <= h;i ++)
            for(int j = 1;j <= w;j ++)
               if(map[i][j] == '#' && hash[i][j] == 0)
               {
                  sum ++;
                  a = i;b = j;
                  BFS();
                }
        printf("%d\n",sum);
    }
}