uva 592 Island of Logic (收索)
Island of Logic |
The Island of Logic has three kinds of inhabitants: divine beings that always tell the truth, evil beings that always lie, and human beings that are truthful during the day and lie at night. Every inhabitant recognizes the type of every other inhabitant.
A social scientist wants to visit the island. Because he is not able to distinguish the three kinds of beings only from their looks, he asks you to provide a communication analyzer that deduces facts from conversations among inhabitants. The interesting facts are whether it is day or night and what kind of beings the speakers are.
Input
The input file contains several descriptions of conversations. Each description starts with an integer
n
, the number of statements in the conversation. The following
n
lines each contain one statement by an inhabitant. Every statement line begins with the speaker's name, one of the capital letters
A
,
B
,
C
,
D
,
E
, followed by a colon `
:
'. Next is one of the following kinds of statements:
I am [not] ( divine | human | evil | lying ).
- X
is [not] ( divine | human | evil | lying ).
It is ( day | night ).
Square brackets [] mean that the word in the brackets may or may not appear, round brackets () mean that exactly one of the alternatives separated by |
must appear. X stands for some name from A, B, C, D, E. There will be no two consecutive spaces in any statement line, and at most 50 statements in a conversation.
The input is terminated by a test case starting with n = 0.
Output
For each conversation, first output the number of the conversation in the format shown in the sample output. Then print ``
This is impossible.
'', if the conversation cannot happen according to the rules or ``
No facts are deducible.
'', if no facts can be deduced. Otherwise print all the facts that can be deduced. Deduced facts should be printed using the following formats:
- X
is ( divine | human | evil ).
It is ( day | night ).
X is to be replaced by a capital letter speaker name. Facts about inhabitants must be given first (in alphabetical order), then it may be stated whether it is day or night.
The output for each conversation must be followed by a single blank line.
Sample Input
1
A: I am divine.
1
A: I am lying.
1
A: I am evil.
3
A: B is human.
B: A is evil.
A: B is evil.
0
Sample Output
Conversation #1
No facts are deducible. Conversation #2
This is impossible. Conversation #3
A is human.
It is night. Conversation #4
A is evil.
B is divine.
题目大意:一个逻辑岛上住有人类,恶魔和神,恶魔总是说谎话,人在白天说真话,在晚上说假话,神一直说真话,现在给出一些人说的话,从中判断并且输出可以确定的信息。
解题思路:将所有语句存成结构体的形式,只需要记录说话人,被说人,说话肯否定(not),和对象。处理方法就是将所有情况列举出来,然后将语句进行判断。如果出现两组满足,就将两组中相同的保留(这些是可以确定的)
注意:即使没有人类被确定,但是白天黑夜已经被分出来了,也要输出。
提供一些数据:
6
A: B is human.
A: B is evil.
B: A is human.
C: A is not lying.
B: C is not human.
D: E is not lying.
4
A: I am human.
A: It is night.
B: I am human.
B: It is day.
3
A: I am human.
B: I am human.
A: B is lying.
3
A: I am divine.
B: A is not lying.
A: B is lying.
3
A: I am divine.
B: A is lying.
A: B is lying.
5
A: B is human.
A: B is evil.
B: A is evil.
C: A is not lying.
B: It is day.
5
C: A is not lying.
A: B is human.
A: B is evil.
B: A is evil.
B: It is day.
1
A: A is not lying.
1
A: A is lying.
2
E: E is evil.
E: E is divine.
7
A: It is night.
B: It is day.
C: I am human.
E: C is human.
C: E is divine.
A: B is lying.
B: C is evil.
0
*********************************************************************************
Conversation #1
A is human.
B is divine.
C is evil.
It is night.
Conversation #2
A is evil.
B is human.
It is day.
Conversation #3
It is day.
Conversation #4
This is impossible.
Conversation #5
No facts are deducible.
Conversation #6
A is evil.
B is divine.
C is evil.
It is day.
Conversation #7
A is evil.
B is divine.
C is evil.
It is day.
Conversation #8
No facts are deducible.
Conversation #9
This is impossible.
Conversation #10
E is human.
It is night.
Conversation #11
A is evil.
C is evil.
E is evil.
It is day.
********************************************************************************
#include<iostream>
#include<string.h>
using namespace std; #define N 55
#define M 10 int people[M];
int brith[M];
int n;
int ok;
int FP;
int now[M];
int yes[M]; struct say{
char talk;
char name;
int bo;
int sex;
}; void thesome()
{
for(int i = 0; i < 6; i++)
if(now[i] != people[i])
yes[i] = 1;
} int ture_say(say f)
{
if(f.sex != -1)
{
if(f.name == 'T')
{
if(f.bo && f.sex - 4 == people[0])
return 1;
else if(!f.bo && f.sex - 4 != people[0])
return 1;
else
return 0;
}
else
{
int id = f.name - 'A' + 1;
if(f.bo && f.sex == people[id])
return 1;
else if(!f.bo && f.sex != people[id])
return 1;
else
return 0;
}
}
else
{
int id = f.name - 'A' + 1; if(people[id] == 3 && !f.bo)
return 1;
else if (people[id] == 1 && f.bo)
return 1;
else if(people[id] == 2)
{
if(people[0] && !f.bo)
return 1;
else if(!people[0] && f.bo)
return 1;
else
return 0;
}
else
return 0;
}
} int write(char str[])
{
if(strncmp(str, "divine.", 7) == 0)
return 3;
else if(strncmp(str, "human.", 6) == 0)
return 2;
else if(strncmp(str, "evil.", 5) == 0)
return 1;
else if(strncmp(str, "day.", 4) == 0)
return 5;
else if( strncmp(str, "night.", 6) == 0)
return 4;
else if( strncmp(str, "lying.", 6) ==0)
return -1;
} void read(say tem[])
{
char str[M];
int id; for(int i = 0; i < n; i++)
{
cin >> str;
tem[i].talk = str[0];
id = str[0] - 'A' + 1;
brith[id] = 1; cin >> str;
if(strcmp(str, "It") == 0)
tem[i].name = 'T';
else if(str[0] == 'I')
tem[i].name = tem[i].talk;
else
{
tem[i].name = str[0];
id = str[0] - 'A' + 1;
brith[id] = 1;
} cin >> str;
cin >> str;
if(strcmp(str, "not"))
{
tem[i].bo = 1;
tem[i].sex = write(str);
}
else
{
tem[i].bo = 0;
cin >> str;
tem[i].sex = write(str);
} // if(tem[i].sex == -1 && tem[i].talk == tem[i].name && tem[i].bo)
// FP = 1;
}
} int judge(say tem[])
{
for(int i = 0; i < n; i++)
{
int id = tem[i].talk - 'A' + 1;
if(people[id] == 3)
{
if(ture_say(tem[i]))
continue;
else
return 0;
}
else if(people[id] == 2)
{
int f = ture_say(tem[i]);
if( (people[0] && f) || (!people[0] && !f) )
continue;
else
return 0;
}
else if(people[id] == 1)
{
if(!ture_say(tem[i]))
continue;
else
return 0;
}
}
return 1;
} void build(int k, say tem[])
{
if(k < 6)
{
if(brith[k])
{
for(people[k] = 1; people[k] < 4; people[k]++)
build(k + 1, tem);
}
else
build(k + 1, tem);
}
else if(judge(tem))
{
// cout << k << endl;
if(ok >= 1)
thesome();
else
for(int i = 0; i < 6; i++)
now[i] = people[i];
ok++;
} } int main()
{
int t = 1;
while(cin >> n, n)
{
// Init.
memset(people, 0, sizeof(people));
memset(brith, 0, sizeof(brith));
ok = 0;
say tem[N];
// FP = 0;
memset(yes, 0,sizeof(yes)); // ReadZZg.
read(tem); cout << "Conversation #" << t++ << endl;
// if(FP)
// cout << "This is impossible." << endl;
// else
// {
for(people[0] = 0; people[0] < 2; people[0]++)
build(1, tem); if(ok == 0)
cout << "This is impossible." << endl;
else
{
//*
int oi = 0;
for(int i = 1; i < 6; i++)
{
if(!yes[i] && now[i])
{
oi++;
char c = 'A' + i - 1;
if(now[i] == 3)
cout << c << " is divine." << endl;
else if(now[i] == 2)
cout << c << " is human." << endl;
else if(now[i] == 1)
cout << c << " is evil." << endl;
}
} if(oi == 0 && yes[0])
cout << "No facts are deducible." << endl;
if(now[0] && !yes[0])
cout << "It is day." << endl;
else if(!yes[0])
cout << "It is night." << endl;
// }
}
cout << endl;
}
return 0;}
uva 592 Island of Logic (收索)的更多相关文章
- 20170712 SQL Server 日志文件收索
-- 1 日志文件增长过快,未进行任务计划截断备份 造成文件过大199G 左右,而可用空间不足8% -- 2 日志备份之前,需要一次完整备份 再进行截断备份 出现可用空间99% 此时可以选择收索数据库 ...
- 剑指offer-第四章解决面试题思路(二叉收索树和双向链表)
题目:输入一个二叉收索树,将二叉搜索树转换成排序的双向链表.要求不能创建节点,只能将链表中的指针进行改变. 将复杂的问题简单化:思路:二叉收索树,本身是一个排序结构,中序遍历二叉收索树就可以得到一组排 ...
- 【转载】快速收索并更新sid 方法
利用Google的搜索功能,可以获得不少SAS各个版本的SID号,试过之后你会异常惊喜.1.打开谷歌: http://google.com.hk2.输入或复制这个段文字:"SID_heade ...
- Combox 实现百度收索框效果
标题中所谓百度收缩框效果,就是在输入数据的时候,自动提示,来张图就明白了: 用Combox来实现这个功能只是需要设置三个A开头的属性就OK了:AutoCompleteSource.AutoComple ...
- Uva592 Island of Logic
题意:神人鬼三个种族,神只说真话,鬼只说假话,人白天说真话,晚上说假话.根据对话内容区分种族和白天黑夜. 最多有A, B, C, D, E五个人 算法:枚举A, B, C, D, E的种族情况和 ...
- 企业常用的站内收索、QQ群、在线客服
<div class="toplinks"> <form target="_blank"> ...
- bootstrapTable 应用小例(收索)
<script src="/plugins/My97DatePicker/WdatePicker.js"></script> <!-- Content ...
- 自定义收索View
1 .h文件 @interface SearchNavView : UIView @property (nonatomic, copy) void(^cancleBtnBlock)(void); @p ...
- UVA题目分类
题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...
随机推荐
- 详细分析Java中断机制(转)
1. 引言 当我们点击某个杀毒软件的取消按钮来停止查杀病毒时,当我们在控制台敲入quit命令以结束某个后台服务时……都需要通过一个线程去取消另一个线程正在执行的任务.Java没有提供一种安全直接的方法 ...
- JPA @PersistenceContext和@Transactional Annotation
JPA(Java Persistence API )也就是说,java存储数据API,它提供的接口更方便的存储数据,当然,经过一些复杂的,并需要使用查询操作Java Persistence query ...
- Duanxx的C++学习: const指针具体解释
Const指的是一个编译时的常量. keywordconst使得代码能够确定一个变量能否够被改动. 使用了const后,能够防止对变量或者指针的改动:更重要的是,const的引用能够防止对所引用的对象 ...
- Matlab绘图-很详细,很全面
强大的绘图功能是Matlab的特点之一,Matlab提供了一系列的绘图函数,用户不需要过多的考虑绘图的细节,只需要给出一些基本参数就能得到 所需图形,这类函数称为高层绘图函数.此外,Matlab还提供 ...
- crm采用soap删除记录
//抽样 function demo() { //操作记录id var targetId = "A8A46444-BA10-E411-8A04-00155D002F02&qu ...
- C# Windows Phone 8 WP8 开发,将WebClient的DownloadStringCompleted事件改成非同步的awiat方法。
原文:C# Windows Phone 8 WP8 开发,将WebClient的DownloadStringCompleted事件改成非同步的awiat方法. 一般我们在撰写Windows Phone ...
- WPF中对三维模型的控制
原文:WPF中对三维模型的控制 (以下选自南开大学出版社出版的<WPF和Silverlight教程>) 3Dmax中的建模模型可以导出为obj文件格式,将此文件导入WPF项目中,由WPF完 ...
- [转载][NAS] 使用win8的“存储池”功能~
之前自己用DQ77KB搭建一个小存储系统(帖子链接:http://www.chiphell.com/thread-567753-1-1.html),一直使用intel主板带的软RAID功能构建RAID ...
- 乐在其中设计模式(C#) - 提供者模式(Provider Pattern)
原文:乐在其中设计模式(C#) - 提供者模式(Provider Pattern) [索引页][源码下载] 乐在其中设计模式(C#) - 提供者模式(Provider Pattern) 作者:weba ...
- 乐在其中设计模式(C#) - 代理模式(Proxy Pattern)
原文:乐在其中设计模式(C#) - 代理模式(Proxy Pattern) [索引页][源码下载] 乐在其中设计模式(C#) - 代理模式(Proxy Pattern) 作者:webabcd 介绍 为 ...