Codeforces Round #396 D

Mahmoud and a Dictionary

题意：一个字典有n个单词，有m个关系表示单词之间是同义还是反义，如果关系可行输出YES如果与前面可行的关系有冲突输出NO，最后k个询问2个单词之间的关系，同义输出1，反义输出2，否则输出3

思路：还是带权或者拆点并查集，如果是有冲突的关系那么不加入并查集

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

map<string,int> M;
string s,s0;
int n,m,q,pre[N<<2];
void init(int n){
    for(int i=0; i<+2*n; ++i) pre[i]=i;
}
int finds(int x){
    return pre[x]=pre[x]==x?x:finds(pre[x]);
}
void unions(int x, int y){
    int fx=finds(x), fy=finds(y);
    pre[fy]=fx;
}
int main(){
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>m>>q;
    init(n);
    for(int i=1; i<=n; ++i){
        cin>>s;
        M[s]=i;
    }
    int t;
    for(int i=1;  i<=m; ++i){
        cin>>t>>s>>s0;
        int x=M[s], y=M[s0];
        if(t==1){
            if(finds(x)==finds(y+n) || finds(y)==finds(x+n)){
                cout<<"NO
";
                continue;
            }
            else{
                cout<<"YES
";
                unions(x,y);
                unions(x+n,y+n);
            }
        }
        else{
            if(finds(x)==finds(y) || finds(x+n)==finds(y+n)){
                cout<<"NO
";
                continue;
            }
            else{
                cout<<"YES
";
                unions(x,y+n);
                unions(y,x+n);
            }
        }
    }
    for(int i=1; i<=q; ++i){
        cin>>s>>s0;
        int x=M[s], y=M[s0];
        if(finds(x)==finds(y)) cout<<"1
";
        else if(finds(x+n)==finds(y)) cout<<"2
";
        else cout<<"3
";
    }
    return 0;
}