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. HibernateReview Day2–Hibernate体系结构

    本文摘自 李刚 著 <Java EE企业应用实战> 现在我们知道了一个概念Hibernate Session,只有处于Session管理下的POJO才具有持久化操作能力.当应用程序对于处于 ...

  2. NSPredicate的用法

    一般来说这种情况还是蛮多的,比如你从文件中读入了一个array1,然后想把程序中的一个array2中符合array1中内容的元素过滤出来. 正 常傻瓜一点就是两个for循环,一个一个进行比较,这样效率 ...

  3. ecshop2.7.3怎么自动清除缓存

    1.在ecs_shop_config表中插入一条数据 进入ECSHOP后台-数据库管理-SQL查询 复制下面SQL,粘贴到里面执行.注意这时是默认表前缀ecs_,如果你的修改过要和你的统一了. INS ...

  4. android应用开发--------------看RadioGroup源代码,写相似单选选项卡的集成控件(如底部导航,tab等等)

    博客为 有时个哥 原创.如需转载请标明出处:http://blog.csdn.net/ls703/article/details/46694967 watermark/2/text/aHR0cDovL ...

  5. 【Java基础】异常的简单分类与处理

    Java中所有的异常都继承自Throwable类,Throwable类的已知子类有Error和Exception. Error是指系统出现的错误,这种错误出现的时候,我们的程序无能为力,所以不需要进行 ...

  6. current online redo logfile 丢失的处理方法

    昨天做了rm -rf操作后的恢复演练,并且是在没有不论什么备份的情况下.今天在做破坏性操作前,做了个rman全备,然后在线删除所有数据库文件,包含控制文件,数据文件,在线日志文件,归档文件等.来看看有 ...

  7. WinForm 国际化开发一例

    1.新建一个WinForm程序(在中文版的Windows下),如下,添加1个Button和1个Label 设置Form1的localizable属性为True(Form1的properties里的De ...

  8. rsync+inotify实现server实时备份

    inotify实现对文件夹下文件进行监听的原理: inotify集成到内核中,通过内核提供的接口.使用inotify作为第三方的软件对文件夹变化进行监控. inotifywait命令能够对文件夹中的文 ...

  9. C++笔试面试总结

    手游广州某公司书面今天接受采访时.刚进去中午1中场休息.他们公司谁刚刚醒来,一个冷漠打牌,然后去上班.瞬间,这些公司有没有什么好印象,压抑. 接着快2点的时候.发了一份笔试题.大部分题目均在网上的&l ...

  10. C++11:强类型枚举(enum)

    // C++11之前的enum类型是继承C的,不温不火: // C++11对enum动刀了,加强了类型检查,推出强类型enum类型,眼前一亮 // 使用过QT 的都知道,早就应该这么做了,用的非常爽! ...