題目連接：http://codeforces.com/contest/766/PRoblem/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
—————————————————————————————-. 題目大意: 就是有n個單詞,有m個關系,q個查詢, 關系是兩個單詞是同義詞還是反義詞,對于當前的這條關系如果與前述關系不符則忽略, 查詢是兩個單詞之間的關系是同義詞還是反義詞還是未知.
解題思路: 很好想到并查集來維護同義詞,然后用另一個數組怎樣的表示兩堆之間的反義詞關系,
最開始是想對每堆進行染色,兩個堆中一個染成+n,另一個染成-n,加和為零表示敵對關系,然后可以在維護一個并查集或者怎樣 ,但是最后GG了
看了題解 發現用另一個數組來維護本堆的反義詞的堆是哪一個，真是奧妙重重。666
這樣的話 這個題目就簡單清晰多了，直接維護就行了。
注意對于m條關系的判定時的細節就好了，詳見代碼。
借(chao)鑒(xi)這里<–戳戳戳
附本題代碼 —————————————————————————————-.
int n,m,q,op;string str,str1,str2;map<string,int>mmp;int pre[N],opp[N];int findi(int x){ int r = x; while(r != pre[r]) r=pre[r]; int i=x,j; while(i!=j){ j=pre[i]; pre[i]=r; i=j; } return r;}void init(){ mmp.clear(); Rep(i,0,n) pre[i]=i,opp[i]=0;}int main(){ while(cin>>n>>m>>q){ init(); Rep(i,1,n) cin>>str,mmp[str]=i; Rep(i,1,m){ cin>>op>>str1>>str2; int x = mmp[str1],y = mmp[str2]; int u = findi(x),v = findi(y); int uu= findi(opp[u]),vv= findi(opp[v]); if(op==1){ if(uu!=v){ puts("YES"); pre[u]=v; if(uu&&vv) pre[uu]=vv; if(!vv) opp[v]=uu; } else puts("NO"); } else { if(u!=v) { puts("YES"); if(uu) pre[v]=uu; else opp[u]=v; if(vv) pre[u]=vv; else opp[v]=u; } else puts("NO"); } } Rep(i,1,q){ cin>>str1>>str2; int x = mmp[str1],y = mmp[str2]; int u = findi(x),v = findi(y); int uu= findi(opp[u]),vv= findi(opp[v]); if (u==v) puts("1"); else { if (uu==v||vv==u) puts("2"); else puts("3"); } } } return 0;}新聞熱點
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