杭电2952 Counting Sheep（简单BFS过）

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2696    Accepted Submission(s): 1799

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
   
   
   
   

 

题目大意：判断联通块的个数（#）

题目很简单，这里直接上代码：

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
struct zuobiao
{
    int x,y;
}now,nex;
char a[122][122];
int vis[121][121];
int fx[4]={0,0,-1,1};
int fy[4]={-1,1,0,0};
int output;
int n,m;
void bfs(int x,int y)
{
    output++;
    now.x=x;
    now.y=y;
    queue<zuobiao >s;
    s.push(now);
    vis[x][y]=1;
    while(!s.empty())
    {
        now=s.front();
        s.pop();
        for(int i=0;i<4;i++)
        {
            nex.x=now.x+fx[i];
            nex.y=now.y+fy[i];
            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&a[nex.x][nex.y]=='#'&&vis[nex.x][nex.y]==0)
            {
                vis[nex.x][nex.y]=1;
                s.push(nex);
            }
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        scanf("%s",a[i]);
        output=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='#'&&vis[i][j]==0)//如果当前是#，并且这个点没有和别的#连通过。进入bfs
                {
                    bfs(i,j);
                }
            }
        }
        printf("%d\n",output);
    }
}