HDU--2952 -- Counting Sheep [广搜]

 

Counting Sheep

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1706    Accepted Submission(s): 1104

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
  
  

 

 

Code:

 

人家都用深搜, 为了练广搜,  哼哼!

可是万恶的Memory Limit Exceeded ! 是Memory ! MLE!  KILL IT !

找到#号的地方把这一块的#号都标志, 原理跟深搜一样一样的~

 

#include<stdio.h>
#include<queue>
#include<string.h>
#include<iostream>
using namespace std;

typedef struct point
{
    int x,y;
}point;

char sign[100][100];
int n,m;
int dir[4][2] = {-1,0,0,1,1,0,0,-1};

void bfs(int a,int b)
{
    queue<point>q;   
    point p,head;
    p.x = a; p.y = b;        
    q.push(p);      
    int i;
    sign[a][b] = 1;   
    while(!q.empty()){      
        head = q.front();
        q.pop(); 
        for(i=0;i<4;i++)//四个方向
        {
            point next;
            next.x = head.x + dir[i][0];
            next.y = head.y + dir[i][1];
            if(next.x>=0 && next.x<n && next.y>=0 && next.y<m && !sign[next.x][next.y]){                                
                q.push(next);
                sign[next.x][next.y] =1;//就是这里!!!改变标志位不能在出队的时候,否则一判断重复就MLE了啊,真是传奇!
            }          
        }    
    }
}

int main()
{
    int t,i,j,count;
    char str[105];
    scanf("%d",&t);
    while(t--)
    {    
        char str[105];
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        {
            scanf("%s",str);
            for(j=0;j<m;j++)
            {
                if(str[j]=='#') sign[i][j] = 0;
                else sign[i][j] = 1;
            }    
        }
        count = 0;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(sign[i][j]==0)
                {
                    bfs(i,j);
                    count ++;
                }               
            }
        }
        printf("%d\n",count);        
    }
    return 0;
}