算法:深搜

题意:让你判断一共有几个羊圈;

思路:像四个方向搜索;

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

代码:

#include <iostream>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
using namespace std;
char a[103][103];
int n,m,k;
int b[4][2]={0,1,0,-1,-1,0,1,0};
void dfs(int x,int y)
{
a[x][y]='.';
for(int i=0;i<4;i++)
{
int dx=x+b[i][0];
int dy=y+b[i][1];
if(dx>=0&&dx<n&&dy>=0&&dy<m&&a[dx][dy]=='#')
dfs(dx,dy);
}
}
int main()
{ int t,i,j;
cin>>t;
while(t--)
{
k=0;
cin>>n>>m;
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
if(a[i][j]=='#')
{ k++;
dfs(i,j);
}
}
cout<<k<<endl;
}
return 0;
}

hdu Counting Sheepsuanga的更多相关文章

  1. HDU 4358 Boring counting(莫队+DFS序+离散化)

    Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) ...

  2. HDU 1264 Counting Squares(线段树求面积的并)

    Counting Squares Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  3. 后缀数组 --- HDU 3518 Boring counting

    Boring counting Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=3518 Mean: 给你一个字符串,求:至少出 ...

  4. hdu 5862 Counting Intersections

    传送门:hdu 5862 Counting Intersections 题意:对于平行于坐标轴的n条线段,求两两相交的线段对有多少个,包括十,T型 官方题解:由于数据限制,只有竖向与横向的线段才会产生 ...

  5. HDU 5862 Counting Intersections (树状数组)

    Counting Intersections 题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Description Given ...

  6. HDU 5952 Counting Cliques 【DFS+剪枝】 (2016ACM/ICPC亚洲区沈阳站)

    Counting Cliques Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) ...

  7. HDU 5862 Counting Intersections(离散化+树状数组)

    HDU 5862 Counting Intersections(离散化+树状数组) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 D ...

  8. HDU 3518 Boring counting

    题目:Boring counting 链接:http://acm.hdu.edu.cn/showproblem.php?pid=3518 题意:给一个字符串,问有多少子串出现过两次以上,重叠不能算两次 ...

  9. HDU 5862 Counting Intersections 扫描线+树状数组

    题目链接: http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Counting Intersections Time Limit: 12000/ ...

随机推荐

  1. C/C++中的拷贝构造函数和赋值构造函数

    代码: #include <iostream> #include <cstdio> using namespace std; class A{ public: A(){ cou ...

  2. Centos6.5使用yum安装Mysql5.7

    想要玩新的东东就要付出代价,我的时间悄悄的都溜走了,说多了都是泪! 实践才是真理! 系统版本:Linux localhost.localdomain 2.6.32-431.el6.x86_64 #1 ...

  3. HTML5 Canvas基础知识

    HTML5画布 1.创建一个画布         <canvas id="myCanvas" width="200" height="100&q ...

  4. 开放GitHub的理由

    越来越多的程序把sourcecode和安装包托管到GitHub上,没有GitHub访问的网络太悲催了... 想通过Chocolatey(windows版的apt-get)装一个ConEmu都无法做到 ...

  5. 同时处理html+js+jquery+css的插件安装(Spket&Aptana插件安装)

    Spket 在线安装方法:Help->Software Updates(或者Install New Software)->Add site Location:http://www.spke ...

  6. 《VIM-Adventures攻略》前言

    本文已转至http://cn.abnerchou.me/2014/03/02/bfdaadb0/ 自从有了计算机,人们就想向其灌输自己的想法. 要想对其输入,自然离不开文本编辑器. 公告:<VI ...

  7. jquery幻灯片--渐变

    网站上为了设计,需要一些幻灯片效果,现在网站有很多插件可以使用. 想要成为以为牛逼的程序员,绝对不允许只会用别人的插件而已,不然你只能是“代码”的搬运工,而不敢做出自己的创新. 首先使用jquery做 ...

  8. Itext 中的文本信息绝对定位

    PdfContentByte pcb = pw.getDirectContent(); pcb.beginText(); pcb.setFontAndSize(bfChinese, 12); pcb. ...

  9. 2015第24周三Spring事务3

    在一个典型的事务处理场景中,有以下几个参与者: Resource Manager(RM) ResourceManager简称RM,它负责存储并管理系统数据资源的状态,比如数据库服务器,JMS消息服务器 ...

  10. 推荐2个小工具 .NET reflector resharper