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…

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…

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>…