counting sheep bfs wyk



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Counting Sheep

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

Total Submission(s) : 8   Accepted Submission(s) : 7

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
ACcode:
#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <limits.h>#include <vector>using namespace std;#define MAX 50005char map[105][105];int cal[105][105];int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int xx,yy;void bfs(int x,int y){    if(x>=0 && x<xx && y>=0 && y<yy)    {        if(map[x][y]=='#' && !cal[x][y])        {            cal[x][y]=1;            for(int i=0; i<4; ++i)                bfs(x+dir[i][0],y+dir[i][1]);        }    }}int main(){    int T;    int i,j,k;    cin>>T;    while(T--)    {        memset(cal,0,sizeof(cal));        scanf("%d%d",&xx,&yy);        for(i=0; i<xx; ++i)        {            getchar();            for(j=0; j<yy; ++j)                //cin>>map[i][j];                scanf("%c",&map[i][j]);        }        int sum=0;        for(i=0; i<xx; ++i)        {            for(j=0; j<yy; ++j)                if(map[i][j]=='#' && cal[i][j]==0)//遇到羊不在一群的羊即没有标记的羊，SUM++                {                    ++sum;                    bfs(i,j);                }        }        cout<<sum<<endl;    }    return 0;}