【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。

 #include<iostream>
 #include<cstring>
 using namespace std;
 const int MAXN = ;
 int Map[MAXN][MAXN] = {};
 int vis[MAXN][MAXN] = {};
 int dfs(int a,int b){
     if(Map[a][b] ==  || vis[a][b] == ) return ;
     vis[a][b] = ;
     //环顾四周
     dfs(a - ,b); //下
     dfs(a + ,b); //上
     dfs(a,b - ); //左
     dfs(a,b + ); //右
     return ;
 }
 int main(){
     int n;
     cin >> n;
     while(n--){
         int a,b,i,j,sum = ;
         memset(Map,,sizeof(Map));
         memset(vis,,sizeof(vis));
         cin >> a >> b;
         for(i = ;i < a;i++){
             for(j = ;j < b;j++){
                 char ac;
                 cin >> ac;
                 if(ac == '#')
                     Map[i][j] = ;
                 else Map[i][j] = ;
             }
         }
         for(i = ;i < a;i++)
         for(j = ;j < b;j++){
             if(Map[i][j] ==  || vis[i][j] == )
                 continue;
             else{

                 dfs(i,j);
                 sum++;
             }
         }
         cout << sum << endl;
     }
     return ;
 }

【题目链接：NYOJ-27】

　　可以说两题完全相似。

 #include<iostream>
 #include<cstring>
 using namespace std;
 const int MAXN = ;
 int Map[MAXN][MAXN] = {};
 int vis[MAXN][MAXN] = {};
 int dfs(int a,int b){
     if(Map[a][b] ==  || vis[a][b] == ) return ;
     vis[a][b] = ;
     //环顾四周
     dfs(a - ,b); //下
     dfs(a + ,b); //上
     dfs(a,b - ); //左
     dfs(a,b + ); //右
     return ;
 }
 int main(){
     int n;
     cin >> n;
     while(n--){
         int a,b,i,j,sum = ;
         memset(Map,,sizeof(Map));
         memset(vis,,sizeof(vis));
         cin >> a >> b;
         for(i = ;i <= a;i++)
         for(j = ;j <= b;j++){
                 cin >> Map[i][j];
         }
         for(i = ;i <= a;i++)
         for(j = ;j <= b;j++){
             if(Map[i][j] ==  || vis[i][j] == )
                 continue;
             else{
                 sum++;
                 dfs(i,j);
             }
         }
         cout << sum << endl;
     }
     return ;
 }