Codeforces Round 396 (Div. 2)
D
Mahmoud and a Dictionary
Codeforces
2000
原题
← CE →
翻译中…

Mahmoud wants to write a new dictionary that contains $*n*$ words and relations between them. There are two types of relations: synonymy (i. e. the two words mean the same) and antonymy (i. e. the two words mean the opposite). From time to time he discovers a new relation between two words.

翻译中…

He know that if two words have a relation between them, then each of them has relations with the words that has relations with the other. For example, if like means love and love is the opposite of hate, then like is also the opposite of hate. One more example: if love is the opposite of hate and hate is the opposite of like, then love means like, and so on.

翻译中…

Sometimes Mahmoud discovers a wrong relation. A wrong relation is a relation that makes two words equal and opposite at the same time. For example if he knows that love means like and like is the opposite of hate, and then he figures out that hate means like, the last relation is absolutely wrong because it makes hate and like opposite and have the same meaning at the same time.

翻译中…

After Mahmoud figured out many relations, he was worried that some of them were wrong so that they will make other relations also wrong, so he decided to tell every relation he figured out to his coder friend Ehab and for every relation he wanted to know is it correct or wrong, basing on the previously discovered relations. If it is wrong he ignores it, and doesn't check with following relations.

翻译中…

After adding all relations, Mahmoud asked Ehab about relations between some words based on the information he had given to him. Ehab is busy making a Codeforces round so he asked you for help.

翻译中…

Input

翻译中…

The first line of input contains three integers $*n*$, $*m*$ and $*q*$ ($2 ≤ *n* ≤ 10^{5}$, $1 ≤ *m*, *q* ≤ 10^{5}$) where $*n*$ is the number of words in the dictionary, $*m*$ is the number of relations Mahmoud figured out and $*q*$ is the number of questions Mahmoud asked after telling all relations.

翻译中…

The second line contains $*n*$ distinct words $*a*_{1}, *a*_{2}, ..., *a*_{*n*}$ consisting of small English letters with length not exceeding $20$, which are the words in the dictionary.

翻译中…

Then $*m*$ lines follow, each of them contains an integer $*t*$ ($1 ≤ *t* ≤ 2$) followed by two different words $*x*_{*i*}$ and $*y*_{*i*}$ which has appeared in the dictionary words. If $*t* = 1$, that means $*x*_{*i*}$ has a synonymy relation with $*y*_{*i*}$, otherwise $*x*_{*i*}$ has an antonymy relation with $*y*_{*i*}$.

翻译中…

Then $*q*$ lines follow, each of them contains two different words which has appeared in the dictionary. That are the pairs of words Mahmoud wants to know the relation between basing on the relations he had discovered.

翻译中…

All words in input contain only lowercase English letters and their lengths don't exceed $20$ characters. In all relations and in all questions the two words are different.

翻译中…

Output

翻译中…

First, print $*m*$ lines, one per each relation. If some relation is wrong (makes two words opposite and have the same meaning at the same time) you should print "NO" (without quotes) and ignore it, otherwise print "YES" (without quotes).

翻译中…

After that print $*q*$ lines, one per each question. If the two words have the same meaning, output $1$. If they are opposites, output $2$. If there is no relation between them, output $3$.

翻译中…

See the samples for better understanding.

翻译中…

样例

Input #1
3 3 4
hate love like
1 love like
2 love hate
1 hate like
love like
love hate
like hate
hate like
Output #1
YES
YES
NO
1
2
2
2
InputCopy OutputCopy
Input #2
8 6 5
hi welcome hello ihateyou goaway dog cat rat
1 hi welcome
1 ihateyou goaway
2 hello ihateyou
2 hi goaway
2 hi hello
1 hi hello
dog cat
dog hi
hi hello
ihateyou goaway
welcome ihateyou
Output #2
YES
YES
YES
YES
NO
YES
3
3
1
1
2
InputCopy OutputCopy
题目信息
题号
766D
时间限制
4 s
内存限制
256 MB
标签
数据结构
DFS 及类似
动态规划
并查集
图论

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。