HDU-2952 Counting Sheep (DFS)

                           Counting Sheep

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 14   Accepted Submission(s) : 12

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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.
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

Source

IDI Open 2009
 1 #include <stdio.h>  
 2 #include <string.h>  
 3  char map[105][105]; 
 4 int f[4][2] = {1,0,-1,0,0,1,0,-1}; //注意定义方向数组。只有四个方向上下左右。
 5 int n,m,s; 
 6 void dfs(int x,int y) 
 7 { 
 8     int i; 
 9     int x1,y1; 
10     for(i = 0;i<4;i++) 
11     { 
12         x1= x+f[i][0]; 
13         y1 = y+f[i][1]; 
14         if(x1<0 || y1<0 || x1>=n || y1>=m || map[x1][y1]!='#') 
15         continue; 
16         map[x1][y1] = '.'; 
17         dfs(x1,y1); 
18     } 
19 } 
20  
21 int main() 
22 { 
23     int t; 
24     scanf("%d",&t); 
25     while(t--) 
26     { 
27         scanf("%d%d",&n,&m); 
28         int i,j,l; 
29         s = 0; 
30         for(i = 0;i<n;i++) 
31         { 
32             scanf("%s",map[i]); 
33         } 
34         for(i = 0;i<n;i++) 
35         { 
36             for(j = 0;j<m;j++) 
37             { 
38                 if(map[i][j] == '#') 
39                 { 
40                     s++; 
41                     map[i][j] = '.'; 
42                     dfs(i,j); 
43                 } 
44             } 
45         } 
46         printf("%d
",s); 
47     } 
48  
49     return 0; 
50 }
原文地址:https://www.cnblogs.com/cancangood/p/3278887.html