模板题-----DFS深度搜索 HDU2952

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2144    Accepted Submission(s): 1412

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

24 4#.#..#.##.##.#.#3 5###.#..#..#.###

 

Sample Output

63
#include<cstdio>#include<iostream>#include<string.h>#define mnum 120using namespace std;char tu[mnum][mnum];int vis[mnum][mnum];const int step_x[4]={0,0,-1,1};const int step_y[4]={1,-1,0,0};int h,w,cnt;void dfs(int i,int j){if(tu[i][j]=='#'&&vis[i][j]==0)//递归函数使用的条件（这里的条件是一旦找到'#'）{vis[i][j]=1; //一定要注意做标记，好像这个叫记忆化搜索！for(int k=0;k<4;k++)  //每个递归点执行什么操作（这里是从递归点向4个方向再递归，直到遇到递归边界为止）{if(i+step_x[k]<1||i+step_x[k]>h)  //递归边界注意把控好，把控不好时数组很容易越界！continue;if(j+step_y[k]<1||j+step_y[k]>w)continue;dfs(i+step_x[k],j+step_y[k]);}}         else        return;//(递归边界) }int main(){int t;scanf("%d",&t);while(t--){memset(vis,0,sizeof(vis));memset(tu,'.',sizeof(tu));scanf("%d%d",&h,&w);getchar();for(int i=1;i<=h;i++){for(int j=1;j<=w;j++){scanf("%c",&tu[i][j]);}getchar();}cnt=0;for(int i=1;i<=h;i++)for(int j=1;j<=w;j++){if(tu[i][j]=='.'||vis[i][j])continue;dfs(i,j);cnt++;}printf("%d\n",cnt);}}