山东省第七届ACM省赛------The Binding of Isaac
The Binding of Isaac
Time Limit: 2000MS Memory limit: 65536K
题目描述
Ok, now I will introduce this game to you...
Isaac is trapped in a maze which has many common rooms…
Like this…There are 9 common rooms on the map.
And there is only one super-secret room. We can’t see it on the map. The super-secret room always has many special items in it. Isaac wants to find it but he doesn’t know where it is.Bob
tells him that the super-secret room is located in an empty place which is adjacent to only one common rooms.
Two rooms are called adjacent only if they share an edge. But there will be many possible places.
Now Isaac wants you to help him to find how many places may be the super-secret room.
输入
Multiple test cases. The first line contains an integer T (T<=3000), indicating the number of test case.
Each test case begins with a line containing two integers N and M (N<=100, M<=100) indicating the number
of rows and columns. N lines follow, “#” represent a common room. “.” represent an empty place.Common rooms
maybe not connect. Don’t worry, Isaac can teleport.
输出
One line per case. The number of places which may be the super-secret room.
示例输入
2 5 3 ..# .## ##. .## ##. 1 1 #
示例输出
8 4
来源
题意
#include"stdio.h" #include"string.h" #include<iostream> using namespace std; char a[105][105]; int main() { int n; cin>>n; while(n--) { int c,b; cin>>c>>b; getchar(); memset(a,0,sizeof(a)); for(int i=1; i<=c; i++) gets(a[i]+1); int s=0; for(int i=0; i<=c+1; i++) for(int j=0; j<=b+1; j++) if(a[i][j]!='#') { int d=0; if(i>0&&a[i-1][j]=='#')d++; if(a[i+1][j]=='#')d++; if(j>0&&a[i][j-1]=='#')d++; if(a[i][j+1]=='#')d++; if(d==1)s++; } printf("%d\n",s); } return 0; }
山东省第七届ACM省赛------The Binding of Isaac的更多相关文章
- 山东省第七届ACM省赛------Memory Leak
Memory Leak Time Limit: 2000MS Memory limit: 131072K 题目描述 Memory Leak is a well-known kind of bug in ...
- 山东省第七届ACM省赛------Reversed Words
Reversed Words Time Limit: 2000MS Memory limit: 131072K 题目描述 Some aliens are learning English. They ...
- 山东省第七届ACM省赛------Triple Nim
Triple Nim Time Limit: 2000MS Memory limit: 65536K 题目描述 Alice and Bob are always playing all kinds o ...
- 山东省第七届ACM省赛------Fibonacci
Fibonacci Time Limit: 2000MS Memory limit: 131072K 题目描述 Fibonacci numbers are well-known as follow: ...
- 山东省第七届ACM省赛------Julyed
Julyed Time Limit: 2000MS Memory limit: 65536K 题目描述 Julyed is preparing for her CET-6. She has N wor ...
- 山东省第七届ACM省赛
ID Title Hint A Julyed 无 B Fibonacci 打表 C Proxy 最短路径 D Swiss-system tournament 归并排序 E The Binding of ...
- 山东省第十届ACM省赛参赛后的学期总结
5.11,5.12两天的济南之旅结束了,我也参加了人生中第一次正式的acm比赛,虽然是以友情队的身份,但是我依旧十分兴奋. 其实一直想写博客来增加自己的能力的,但是一直拖到现在,正赶上老师要求写一份总 ...
- Rectangles(第七届ACM省赛原题+最长上升子序列)
题目链接: http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=1255 描述 Given N (4 <= N <= 100) rec ...
- 山东理工大学第七届ACM校赛-LCM的个数 分类: 比赛 2015-06-26 10:37 18人阅读 评论(0) 收藏
LCM的个数 Time Limit: 1000ms Memory limit: 65536K 有疑问?点这里^_^ 题目描述 对于我们来说求两个数的LCM(最小公倍数)是很容易的事,现在我遇到了 ...
随机推荐
- javascript立即执行函数
javascript和其他编程语言相比比较随意,所以javascript代码中充满各种奇葩的写法,有时雾里看花;当然,能理解各型各色的写法也是对javascript语言特性更进一步的深入理解. ( ...
- 浅谈java抽象类和接口
第一次,写这个,没有把图片放上来,有兴趣的可以点击连接看原文 http://note.youdao.com/noteshare?id=aecbd52b9240f23c0954e8086b848a17 ...
- 浅谈iOS开发中方法延迟执行的几种方式
Method1. performSelector方法 Method2. NSTimer定时器 Method3. NSThread线程的sleep Method4. GCD 公用延迟执行方法 - (vo ...
- Reveal - UI 分析工具
一.安装和简介 a) download url b) Reveal 使用的方法有两种: Static Library Intefration, Dynamic Library Intefration. ...
- flex 弹性布局
采用 flex 布局的元素称为容器,其所有子元素称为项目.常用: 容器上可以设置六个属性:flex-direction.flex-wrap.flex-flow.justify-content ...
- 解决Rational Rose 中 没有 Data modeler 选项的问题
在 Rose 没有 Data modeler 选项的原因是没有将 Data modeler 这块功能勾选上. 解决方案: 菜单栏--Add-Ins--Add-Ins Manager-->找到 ...
- Android界面组件的四种启动方式
Android界面组件启动有四种方式 standard,singleTop,singleTask,singleInstance. standard:每次调用都会都会产生新的组件. singletop: ...
- http之Session&Cookie
百度了一波session与Cookie,我发现这东西远比我想象中更复杂(可能是因为我不明白底层的运行原理).网上也是一堆的关于Session与Cookie区别/联系的文章,然而,我看完了还是一脸懵逼的 ...
- django小结
初始化项目 python manage.py runserver python manage.py runserver 127.0.0.1:8080 python manage.py startapp ...
- Sql Server中暂停命令
Sql Server中暂停几秒再执行后面的命令! -- 语法WAITFOR { DELAY 'time_to_pass' | TIME 'time_to_execute' | [ ( r ...