Mahmoud and a Dictionary codeforces396 div2 D

D. Mahmoud and a Dictionary
time limit per test4 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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: if love 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 means like, 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.

Input
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.

Output
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.

Examples
input
3 3 4
hate love like
1 love like
2 love hate
1 hate like
love like
love hate
like hate
hate like
output
YES
YES
NO
1
2
2
2
input
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
YES
YES
YES
YES
NO
YES
3
3
1
1
2

题意：
给你n个单词,m个关系(两个单词是反义词还是同义词);
然后问你所给的关系里面有没有错的(就是互相抵触了);
最后再给你q个询问,问你两个单词之间的关系是什么;
同义词输出1,反义词输出2,不确定输出3;

带权并查集：不懂的可以看下这道：http://blog.csdn.net/littlewhite520/article/details/53521037
只有两个值 1和 2 两种状态所以%2就行，保持在 0 1，要输出不确定的为3，记住，不在一棵树里面的就是不确定的。

#include <bits/stdc++.h>using namespace std;const int maxn=1e5+7;int f[maxn];int relation[maxn];map <string,int > mp;int find(int x){    if(x!=f[x])    {        int t=f[x];        f[x]=find(f[x]);        relation[x]=(relation[x]+relation[t])%2;    }    return f[x];}int main(){    int n,m,q;    while(cin>>n>>m>>q)    {        string s;        for(int i=0;i<n;i++)        {            cin>>s;            mp[s]=i;        }        for(int i=0;i<n;i++)        {            f[i]=i;            relation[i]=0;        }        string s1;        int v;        while(m--)        {            cin>>v>>s>>s1;            v--;            int f1=mp[s];            int f2=mp[s1];            int t1=find(f1);            int t2=find(f2);            if(t1!=t2)            {                printf("YES\n");                f[t1]=t2;                relation[t1]=(v-relation[f1]+relation[f2]);            }            else            {                int p=(relation[f1]-relation[f2]+2)%2;                if(p==v) puts("YES");                else                    puts("NO");            }            }         while(q--)         {            cin>>s>>s1;            int f1=mp[s];            int f2=mp[s1];            int t1=find(f1);            int t2=find(f2);            if(t1!=t2)            {                puts("3");            }            else            {                int p=(relation[f1]-relation[f2]+2)%2;                printf("%d\n",p+1 );            }         }    }}