hdu Counting Sheepsuanga
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
- 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) ...
- 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) ...
- 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 ...
- 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 ...
- 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 ...
- 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) ...
- 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 ...
- 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 ...
- 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
- Tarjan求极大强连通分量模板
#include<iostream> #include<cstring> #include<cstdio> #include<stack> #inclu ...
- vmware workstation 7.0官方下载安装
https://my.vmware.com/group/vmware/downloads#tab2 这里注册后可以下载到vmware的所有产品,可以下载到免费的vmplayer,以及收费的vmware ...
- android ioctl fuzz,android 本地提权漏洞 android root
目前正在研究android 三方设备驱动 fuzzer , 也就是下图所说的 ioctl fuzzing, 下图是由keen team nforest 大神发布: 欢迎正在研究此方面的人联系我共同交流 ...
- MySQL binlog_rows_query_log_events
当binlog_format=statement的时候进制日志只记录的是SQL语句,当binlog_fromat=row的时候记录的是event,如果想要在row模式的情况下 也记录SQL语句:bin ...
- bat命令大全
一.简单批处理内部命令简介 1.Echo 命令 打开回显或关闭请求回显功能,或显示消息.如果没有任何参数,echo 命令将显示当前回显设置. 语法 echo [{on│off}] [message ...
- Delphi中多线程下使用使用 UniDAC+MSSQL 需要注意的问题(连接前调用CoInitialize)
一般解决方法是在线程开始启用 CoInitialize(nil),线程结束调用 CoUninitialize .如果你使用多种数据库连接,比如三层中经常切换到MSSQL和Oracle,我们只需在判断 ...
- JAVA中的时间操作
java中的时间操作不外乎这四种情况: 1.获取当前时间 2.获取某个时间的某种格式 3.设置时间 4.时间的运算 好,下面就针对这四种情况,一个一个搞定. 一.获取当前时间 有两种方式可以获得,第一 ...
- EXTJS4:在grid中加入合计行
extjs4很方便的实现简单的合计(针对在不分页的情况下): 它效果实现在:Ext.grid.feature.Summary这个类中 Ext.define('TestResult', { extend ...
- 华为手机root 删除一般不用软件 的命令
上个B518海外版的一键root精简 精简了以下这些,不想删除的自己可以在刷机脚本中删除对应行就行了,音量解锁,GPS,搜索键关屏,root,添加钛备份4.0,re管理器,其他框架未改动,稳定性不会变 ...
- 使用layer显示弹出框笔记
$.layer({ area : ['200px','auto'], //控制层宽高.当设置为auto时,意味着采用自适应, 当然,对于宽度,并不推荐这样做.例如:area : ['310px ...