hdu2952――Counting Sheep（DFS&&BFS）

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
   
   
   
   

分析：

四个方向暴搜；

源码：（DFS）

#include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> using namespace std; int m,n; char map[120][120]; int i,j; void bfs(int i,int j) {     if(map[i][j]!='#'||i<0||j<0||i>=m||j>=n)     return ;     else     {         map[i][j]='.';         bfs(i,j+1);         bfs(i,j-1);         bfs(i+1,j);         bfs(i-1,j);     } } int main() {     int tase;     scanf("%d",&tase);     while(tase--)     {         memset(map,0,sizeof(map));         scanf("%d%d",&m,&n);         for(i=0; i<m; i++)             for(j=0; j<n; j++)                 cin>>map[i][j];         int p=0;         for(i=0; i<m; i++)             for(j=0; j<n; j++)             {                 if(map[i][j]=='#')                 {                     bfs(i,j);                     p++;                 }             }             printf("%d\n",p);     }     return 0; } 

源码：（BFS）

#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; int m,n; int i,j; int p; char map[120][120]; int visit[4][2]= {1,0,-1,0,0,1,0,-1}; int nextx,nexty; void bfs(int x,int y) {     for(int i=0; i<4; i++)     {         nextx=x+visit[i][0];         nexty=y+visit[i][1];         if(nextx<0||nexty<0||nextx>=m||nexty>=n||map[nextx][nexty]!='#')             continue;         map[nextx][nexty]='.';         bfs(nextx,nexty);     } } int main() {     int test;     scanf("%d",&test);     while(test--)     {         p=0;         memset(map,0,sizeof(map));         scanf("%d%d",&m,&n);         for(i=0; i<m; i++)             for(j=0; j<n; j++)                 cin>>map[i][j];         for(i=0; i<m; i++)             for(j=0; j<n; j++)             {                 if(map[i][j]=='#')                 {                     map[i][j]=='.';                     bfs(i,j);                     p++;                 }                 else continue;             }         printf("%d\n",p);     }     return 0; } /* 2 4 4 #.#. .#.# #.## .#.# 3 5 ###.# ..#.. #.### */