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[23456], *S = RR + 23333, *T = RR + 23333;
    if(S == T) fread(RR, 1, 23333, stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = 0, w = 1; char c = gc();
    while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
    while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
    return p * w;
}

#define pc(o) *O ++ = o
char WR[30000005], *O = WR;
inline void write(int opt) {
    if(opt == 1) pc('Y'), pc('E'), pc('S'), pc('
');
    else pc('N'), pc('O'), pc('
');
}

#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 = 1; i <= n; i ++) {
        x[i] = read(); y[i] = read(); p[i] = read();
        T[++ tot] = x[i]; T[++ tot] = y[i];
    }
    sort(T + 1, T + tot + 1);
    tot = unique(T + 1, T + tot + 1) - T - 1;
    for(ri i = 1; i <= n; i ++) {
        x[i] = lower_bound(T + 1, T + tot + 1, x[i]) - T;
        y[i] = lower_bound(T + 1, T + tot + 1, y[i]) - T;
    }
    for(ri i = 1; i <= tot; i ++) fa[i] = i;
    for(ri i = 1; i <= n; i ++) {
        int u = find(x[i]), v = find(y[i]);
        if(!p[i]) {
            if(u == v) write(-1);
            else write(1), f[u].insert(v), f[v].insert(u);
        }
        else {
            if(u == v) write(1);
            else if(f[u].count(v)) write(-1);
            else {
                write(1);
                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, 1, O - WR, stdout);
    return 0;
}