Counting Sheep

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

Total Submission(s) : 9   Accepted Submission(s) : 8

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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.

HDU-2952 Counting Sheep(bfs/dfs)_#include

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


思路

连在一起的是一个羊群，让输出羊群的个数。

 

先找到一只羊，然后搜索所有和他属于一个羊群的羊，并把该羊群的所有点标志为已搜索。

第一次搜索结束之后，res++ ，接着，从另一个未被搜过的羊开始找另一个羊群。

......循环

直到找到这片区域。

 

深搜代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int maxn = 100 + 10;
int dir[8] = {1,0,-1,0,0,1,0,-1};
char a[maxn][maxn];
bool visited[maxn][maxn];
int h,w,res;

void dfs(int n, int m){
  
  visited[n][m] = true;
  for (int i = 0; i < 8; i += 2) {
    if ((n+dir[i]) >= 0 && (m+dir[i+1]) >= 0 && (n+dir[i]) < h && (m+dir[i+1]) < w
      && !visited[n+dir[i]][m+dir[i+1]] && a[n+dir[i]][m+dir[i+1]] == '#') {
        dfs(n+dir[i], m+dir[i+1]);
      }
  }
}

int main() {
  int t;
  scanf("%d", &t);
  while(t--) {
    scanf("%d%d", &h, &w);
    memset(a, 0, sizeof(a));
    memset(visited, false, sizeof(visited));
    res = 0;
    for (int i = 0; i < h; i++) {
      scanf("%s", a[i]);
    }
    for (int i = 0; i < h; i++) {
      for (int j = 0; j < w; j++) {
        if (!visited[i][j] && a[i][j] == '#') {
          res++;
          dfs(i, j);
        }
      }
    }
    printf("%d\n", res);
  }
  return 0;
}

广搜代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;


struct node{
  int x;
  int y;
  node(int xx, int yy):x(xx),y(yy){}
};
const int maxn = 100 + 10;
int dir[8] = {1,0,-1,0,0,1,0,-1};
char a[maxn][maxn];
bool visited[maxn][maxn];
int h,w,res;


void bfs(int n, int m){
  queue<node> q;
  q.push(node(n,m));
  res++;
  while(!q.empty()){
    node p = q.front();
    q.pop();
    int x = p.x;
    int y = p.y;
    for (int i = 0; i < 8; i += 2) {
      
      if (x + dir[i] < h &&  y + dir[i+1] < w && (y + dir[i+1]) >= 0 && (x + dir[i]) >= 0
        && !visited[x + dir[i] ][ y + dir[i+1] ] && a[x + dir[i]][y + dir[i+1]] == '#') {
          q.push(node(x + dir[i],y + dir[i+1]));
          visited[x + dir[i]][y + dir[i+1]] = true;
        }
    }
    
    
  }
  
}


int main() {
  int t;
  scanf("%d", &t);
  while(t--) {
    scanf("%d%d", &h, &w);
    memset(a, 0, sizeof(a));
    memset(visited, false, sizeof(visited));
    res = 0;
    for (int i = 0; i < h; i++) {
      scanf("%s", a[i]);
    }
    for (int i = 0; i < h; i++) {
      for (int j = 0; j < w; j++) {
        if (!visited[i][j] && a[i][j] == '#') {
          bfs(i, j);
        }   
      }
    }
    printf("%d\n", res);
  }
  return 0;
}