hdu 2952 田字格四个方向的深度优先搜索 计算连通分量

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Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2665    Accepted Submission(s): 1768

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

分析：

田字格的四个方向搜索

#include<iostream>#include<stdlib.h>#include<stdio.h>#include<stack>#include<string>#include<algorithm>using namespace std;#define maxn 105int dir[4][2] = {{-1 , 0} , {1 , 0} , {0 , -1} , {0 , 1}   } ; //按照上下左右的顺序char arc[maxn][maxn] ;int h  ,w ;void dfs(int x , int y){  //  cout << "x = " <<x << "y = " << y << endl ;    int nowx , nowy ;    for(int i = 0 ; i<4 ; i++){        nowx = x + dir[i][0] ;        nowy = y + dir[i][1] ;         if(nowx != -1 && nowy != -1 && arc[nowx][nowy] == '#'  ){          //  cout << " dir i = " << i << endl;            arc[nowx][nowy] = '.' ;            dfs(nowx , nowy) ;          //  cout << "return x = " << x << "y = " << y  << endl;         }    }}void solve(){    int cou = 0 ;    for(int i = 0 ; i < h; i++){        for(int j = 0 ; j< w ; j++){            if(arc[i][j] == '#'){                arc[i][j] = '.' ;                dfs(i , j) ;                cou++ ;            }        }   }   cout << cou << endl;}int main(){    int t;    cin>>t ;    while(t--){        cin>>h >> w ;        for(int i =0 ;  i<h ; i++){            for(int j = 0 ; j<w ; j++ ){                cin >> arc[i][j] ;            }        }        for(int i = 0 ;  i<= h ; i++)            arc[i][w] = '.' ;        for(int i =0 ; i<= w ; i++)            arc[h][i] = '.' ;//        for(int i =0 ;  i<=h ; i++){//            for(int j = 0 ; j<=w ; j++ ){//                printf("arc[%d][%d] = %c " , i  ,j , arc[i][j] );//            }//            cout<<"\n" << endl;//        }        solve() ;    }   return 0 ;}