51nod1515 明辨是非 并查集 + set

一开始想的时候，好像两个并查集就可以做......然后突然懂了什么....

相同的并查集没有问题，不同的就不能并查集了，暴力的来个set就行了.....

合并的时候启发式合并即可做到$O(n \log^2 n)$

如果打$splay$，那么启发式合并可以做到$O(n \log n)$

#include <set>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

extern inline char gc() {
    static char RR[], *S = RR + , *T = RR + ;
    if(S == T) fread(RR, , , stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = , w = ; char c = gc();
    while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
    while(c >= '' && c <= '') p = p *  + c - '', c = gc();
    return p * w;
}

#define pc(o) *O ++ = o
char WR[], *O = WR;
inline void write(int opt) {
    if(opt == ) pc('Y'), pc('E'), pc('S'), pc('\n');
    else pc('N'), pc('O'), pc('\n');
}

#define ri register int
#define sid 200050

set <int> f[sid];
int T[sid], n, tot;
int x[sid], y[sid], p[sid], fa[sid];

inline int find(int o) {
    if(fa[o] == o) return o;
    return fa[o] = find(fa[o]);
}

int main() {
    n = read();
    for(ri i = ; i <= n; i ++) {
        x[i] = read(); y[i] = read(); p[i] = read();
        T[++ tot] = x[i]; T[++ tot] = y[i];
    }
    sort(T + , T + tot + );
    tot = unique(T + , T + tot + ) - T - ;
    for(ri i = ; i <= n; i ++) {
        x[i] = lower_bound(T + , T + tot + , x[i]) - T;
        y[i] = lower_bound(T + , T + tot + , y[i]) - T;
    }
    for(ri i = ; i <= tot; i ++) fa[i] = i;
    for(ri i = ; i <= n; i ++) {
        int u = find(x[i]), v = find(y[i]);
        if(!p[i]) {
            if(u == v) write(-);
            else write(), f[u].insert(v), f[v].insert(u);
        }
        else {
            if(u == v) write();
            else if(f[u].count(v)) write(-);
            else {
                write();
                if(f[u].size() > f[v].size()) swap(u, v); fa[u] = v;
                for(set <int> :: iterator it = f[u].begin(); it != f[u].end(); it ++)
                { int w = *it; f[w].insert(v); f[v].insert(w); }
            }
        }
    }
    fwrite(WR, , O - WR, stdout);
    return ;
}