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

24 4#.#..#.##.##.#.#3 5###.#..#..#.###

 

Sample Output

63

 

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);    }}