standard output
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.
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: iflove is the opposite of hate and hate is the opposite of like, then love means like, and so on.
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 meanslike, the last relation is absolutely wrong because it makes hate and like opposite and have the same meaning at the same time.
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.
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.
The first line of input contains three integers n, m and q (2 ≤ n ≤ 105, 1 ≤ m, q ≤ 105) 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.
The second line contains n distinct words a1, a2, ..., an consisting of small English letters with length not exceeding 20, which are the words in the dictionary.
Then m lines follow, each of them contains an integer t (1 ≤ t ≤ 2) followed by two different words xi and yi which has appeared in the dictionary words. If t = 1, that means xi has a synonymy relation with yi, otherwise xi has an antonymy relation with yi.
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.
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.
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).
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.
See the samples for better understanding.
3 3 4
hate love like
1 love like
2 love hate
1 hate like
love like
love hate
like hate
hate like
YES
YES
NO
1
2
2
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
YES
YES
YES
YES
NO
YES
3
3
1
1
2
题目链接：http://codeforces.com/contest/766/problem/D
题意：有n个串，m个关系，q个询问。告诉两个串是同义还是反义。与之前的关系矛盾输出NO，否则关系成立。输出每次询问的两个串
之间的关系，同义输出1，反义输出2，不确定输出3。
思路：并查集。2*i表示这个串，2*i-1表示这个串的反义。如果i和j（i和j表示串的标号）同义，则2*i与2*j，2*i-1与2*j-1同义。如果i和j反义，则2*i与
2*j-1,2*j与2*i-1同义。
代码：
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
const int MAXN=1e6+;
int pa[MAXN];
map<string,int>sign;
int findset(int x)
{
return pa[x]!=x?pa[x]=findset(pa[x]):x;
}
void uni(int s,int e)
{
int pas=findset(s),pae=findset(e);
if(pas!=pae) pa[pas]=pae;
}
int main()
{
int n,m,q;
scanf("%d%d%d",&n,&m,&q);
for(int i=; i<=*n; i++) pa[i]=i;
for(int i=; i<=n; i++)
{
string ch;
cin>>ch;
sign[ch]=i;
}
string s,e,pas,pae;
for(int i=; i<m; i++)
{
int t;
int ans=;
scanf("%d",&t);
cin>>s>>e;
int id1=sign[s],id2=sign[e];
if(t==)
{
if(findset(*id1)==findset(*id2-)) cout<<"NO"<<endl;
else if(findset(*id2)==findset(*id1-)) cout<<"NO"<<endl;
else
{
cout<<"YES"<<endl;
uni(*id1,*id2);
uni(*id1-,*id2-);
}
}
else
{
if(findset(*id1)==findset(*id2)) cout<<"NO"<<endl;
else if(findset(*id1-)==findset(*id2-)) cout<<"NO"<<endl;
else
{
cout<<"YES"<<endl;
uni(*id1,*id2-);
uni(*id1-,*id2);
}
}
}
for(int i=; i<q; i++)
{
cin>>s>>e;
int id1=sign[s],id2=sign[e];
if(findset(*id1)==findset(*id2)) cout<<<<endl;
else if(findset(*id1-)==findset(*id2-)) cout<<<<endl;
else if(findset(*id1)==findset(*id2-)) cout<<<<endl;
else if(findset(*id2)==findset(*id1)) cout<<<<endl;
else cout<<<<endl;
}
return ;
}
并查集