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 discov…

D. Mahmoud and a Dictionary time limit per test:4 seconds memory limit per test:256 megabytes input:standard input output: standard output Mahmoud wants to write a new dictionary that contains n words and relations between them. There are two types o…

D. Mahmoud and a Dictionary 题目连接: http://codeforces.com/contest/766/problem/D Description 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 t…

D. Mahmoud and a Dictionary time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud wants to write a new dictionary that contains n words and relations between them. There are two types of…

地址:http://codeforces.com/contest/766/problem/D 题目: D. Mahmoud and a Dictionary time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud wants to write a new dictionary that contains n words…

Mahmoud and a Dictionary time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud wants to write a new dictionary that contains n words and relations between them. There are two types of re…

D. Mahmoud and a Dictionary time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud wants to write a new dictionary that contains n words and relations between them. There are two types of…

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…

题意:给出n个单词,m条关系,q个询问,每个对应关系有,a和b是同义词,a和b是反义词,如果对应关系无法成立就输出no,并且忽视这个关系,如果可以成立则加入这个约束,并且输出yes.每次询问两个单词的关系,1,同义词,2,反义词,3,不确定 题解:这题思路比较奇特,开辟2*n的并查集的空间,第i+n代表i的反义词所在的树,初始为i+n,也就是说i+n代表i的反义词 #include<bits/stdc++.h> using namespace std; #define ll long long…

并查集. 将每一个物品拆成两个,两个意义相反,然后并查集即可. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include<set>…