【DFS深搜初步】HDOJ-2952 Counting Sheep、NYOJ-27 水池数目

【题目链接：HDOJ-2952】

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2476    Accepted Submission(s): 1621

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

【思路】

　　深搜：就是把每种可能都枚举出来，直到找到符合条件的可能。

 1 #include<iostream>

 2 #include<cstring>

 3 using namespace std;

 4 const int MAXN = 105;

 5 int Map[MAXN][MAXN] = {0};

 6 int vis[MAXN][MAXN] = {0};

 7 int dfs(int a,int b){

 8     if(Map[a][b] == 0 || vis[a][b] == 1) return 0;

 9     vis[a][b] = 1;

10     //环顾四周 

11     dfs(a - 1,b); //下

12     dfs(a + 1,b); //上

13     dfs(a,b - 1); //左

14     dfs(a,b + 1); //右 

15     return 0;

16 }

17 int main(){

18     int n;

19     cin >> n;

20     while(n--){

21         int a,b,i,j,sum = 0;

22         memset(Map,0,sizeof(Map));

23         memset(vis,0,sizeof(vis));

24         cin >> a >> b;

25         for(i = 0;i < a;i++){      

26             for(j = 0;j < b;j++){

27                 char ac;

28                 cin >> ac; 

29                 if(ac == '#')

30                     Map[i][j] = 1;

31                 else Map[i][j] = 0;                                    

32             }            

33         }                

34         for(i = 0;i < a;i++)

35         for(j = 0;j < b;j++){

36             if(Map[i][j] == 0 || vis[i][j] == 1)

37                 continue;

38             else{

39                 

40                 dfs(i,j);

41                 sum++;

42             }

43         }

44         cout << sum << endl; 

45     }

46     return 0;

47 }

【题目链接：NYOJ-27】

　　可以说两题完全相似。

 1  

 2 #include<iostream>

 3 #include<cstring>

 4 using namespace std;

 5 const int MAXN = 105;

 6 int Map[MAXN][MAXN] = {0};

 7 int vis[MAXN][MAXN] = {0};

 8 int dfs(int a,int b){

 9     if(Map[a][b] == 0 || vis[a][b] == 1) return 0;

10     vis[a][b] = 1;

11     //环顾四周 

12     dfs(a - 1,b); //下

13     dfs(a + 1,b); //上

14     dfs(a,b - 1); //左

15     dfs(a,b + 1); //右 

16     return 0;

17 }

18 int main(){

19     int n;

20     cin >> n;

21     while(n--){

22         int a,b,i,j,sum = 0;

23         memset(Map,0,sizeof(Map));

24         memset(vis,0,sizeof(vis));

25         cin >> a >> b;

26         for(i = 1;i <= a;i++)

27         for(j = 1;j <= b;j++){

28                 cin >> Map[i][j];

29         }

30         for(i = 1;i <= a;i++)

31         for(j = 1;j <= b;j++){

32             if(Map[i][j] == 0 || vis[i][j] == 1)

33                 continue;

34             else{

35                 sum++;

36                 dfs(i,j);

37             }

38         }

39         cout << sum << endl; 

40     }

41     return 0;

42 }