【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); //
12     dfs(a + 1,b); //
13     dfs(a,b - 1); //
14     dfs(a,b + 1); //
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); //
13     dfs(a + 1,b); //
14     dfs(a,b - 1); //
15     dfs(a,b + 1); //
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 }        
 
原文地址:https://www.cnblogs.com/zhengbin/p/4482992.html