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 ABCDE. 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

  1. 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 ...

  2. 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 ...

  3. 【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 ...

  4. 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 ...

  5. 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 ...

  6. 企业常用的站内收索、QQ群、在线客服

    <div class="toplinks">            <form target="_blank">             ...

  7. bootstrapTable 应用小例(收索)

    <script src="/plugins/My97DatePicker/WdatePicker.js"></script> <!-- Content ...

  8. 自定义收索View

    1 .h文件 @interface SearchNavView : UIView @property (nonatomic, copy) void(^cancleBtnBlock)(void); @p ...

  9. 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

  1. ORACLE 中的 锁 介绍

    ORACLE 中的 锁 介绍 Oracle数据库支持多个用户同时与数据库进行交互,每个用户都可以同时运行自己的事务,从而也需要对并发访问进行控制.Oracle也是用“锁”的机制来防止各个事务之间的相互 ...

  2. hash表、hash算法

    概念: 散列表(Hash table.也叫哈希表),是依据关键码值(Key value)而直接进行訪问的数据结构. 也就是说,它通过把关键码值映射到表中一个位置来訪问记录,以加快查找的速度.这个映射函 ...

  3. veridata实验例(3)验证veridata发现insert操作不会导致同步

    veridata实验例(3)验证veridata发现insert操作不会导致同步 续接:<veridata实验举例(2)验证表BONUS与表SALGRADE两节点同步情况>,地址:点击打开 ...

  4. QQ旅程server分析01-网关server

    网关server网络进程: s1.连接网关管理器 s2.上报自身信息给网关管理器 s3.从网关管理器获取自己的管理信息 *s4.依照管理信息预分配好须要的资源 s5.网关server打开client监 ...

  5. vim使用(三):.viminfo和.vimrc

    1. viminfo 在vim中操作的行为,vim会自己主动记录下来,保存在 ~/.viminfo 文件里. 这样为了方便下次处理, 如:vim打开文件时,光标会自己主动在上次离开的位置显示. 原来搜 ...

  6. TCP/IP详细说明--滑模、拥塞窗口、慢启动、Negle算法

    TCP的数据流大致能够分为两类,交互数据流与成块的数据流. 交互数据流就是发送控制命令的数据流.比方relogin,telnet.ftp命令等等.成块数据流是用来发送数据的包,网络上大部分的TCP包都 ...

  7. Memento pattern

    21.5 再谈备忘录的封装 备忘录是一个很特殊的对象,只有原发器对它拥有控制的权力,负责人只负责管理,而其他类无法访问到备忘录,因此我们需要对备忘录进行封装. 为了实现对备忘录对象的封装,需要对备忘录 ...

  8. Lucene40SkipListWriter

    多级跳跃表是保存在tim文件里的. tip是term index,tim是term dictionary.记忆方法是,p是pointer因此是term index. 这个类会保存多个level的las ...

  9. DB2 “The transaction log for the database is full” 存在的问题及解决方案

    DB2在执行一个大的insert/update操作的时候报"The transaction log for the database is full.. "错误,查了一下文档是DB ...

  10. 升级 树莓派的gcc/g++编译器到4.8

    首先,更新你的树莓派(避免编译出来版本不对导致不能insert)sudo apt-get update sudo apt-get upgrade sudo apt-get dist-upgrade s ...