CF603EPastoral Oddities
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/*
LCT管子题(说的就是你 水管局长)
首先能得到一个结论, 那就是当且仅当所有联通块都是偶数时存在构造方案
LCT动态加边, 维护最小生成联通块, 用set维护可以删除的边, 假如现在删除后不影响全都是偶数大小的性质 就删除
不清楚link为啥要makeroot两次
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#include<set>
#include<cmath>
#define ll long long
#define M 400010
#define mmp make_pair
using namespace std;
int read() {
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
int n, m, cnt, ln[M], rn[M];
struct LCT {
int fa[M], ch[M][2], sz[M], rev[M], val[M], mx[M], pl[M], zz[M];
#define ls ch[x][0]
#define rs ch[x][1]
void pushup(int x) {
sz[x] = sz[ls] + sz[rs] + zz[x] + 1;
mx[x] = max(mx[ls], mx[rs]);
pl[x] = (mx[x] == mx[ls]) ? pl[ls]: pl[rs];
if(val[x] > mx[x]) mx[x] = val[x], pl[x] = x;
}
bool isr(int x) {
return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
}
void add(int x) {
rev[x] ^= 1;
swap(ls, rs);
}
void pushdown(int x) {
if(rev[x]) {
add(ls), add(rs);
rev[x] = 0;
}
}
void pd(int x) {
if(!isr(x)) pd(fa[x]);
pushdown(x);
}
void rotate(int x) {
int y = fa[x], q = fa[y];
bool dy = (ch[y][1] == x), dz = (ch[q][1] == y);
if(!isr(y)) ch[q][dz] = x;
fa[x] = q;
fa[ch[x][dy ^ 1]] = y;
ch[y][dy] = ch[x][dy ^ 1];
ch[x][dy ^ 1] = y;
fa[y] = x;
pushup(y);
pushup(x);
}
void splay(int x) {
pd(x);
while(!isr(x)) {
int y = fa[x];
if(!isr(y)) {
int q = fa[y];
if((ch[y][1] == x) ^ (ch[q][1] == y)) rotate(x);
else rotate(y);
}
rotate(x);
}
}
void access(int x) {
for(int y = 0; x; y = x, x = fa[x]) splay(x), zz[x] += sz[rs] - sz[y], rs = y, pushup(x);
}
void maker(int x) {
access(x);
splay(x);
add(x);
}
int findr(int x) {
access(x);
splay(x);
while(ls) pushdown(x), x = ls;
return x;
}
void link(int x, int y) {
maker(x);
maker(y);
fa[x] = y;
zz[y] += sz[x];
sz[y] += sz[x];
}
void split(int x, int y) {
maker(x), access(y), splay(y);
}
void cut(int x, int y) {
split(x, y);
if(fa[x] != y && ch[x][1]) return;
fa[x] = ch[y][0] = 0;
pushup(y);
}
int query(int x, int y) {
split(x, y);
return pl[y];
}
int sum(int x) {
maker(x);
return ((sz[x] + 1) >> 1) & 1;
}
} lct;
set<pair<int, int> >st;
set<pair<int, int> >::reverse_iterator it;
int main() {
n = read(), m = read();
cnt = n;
// for(int i = 1; i <= n; i++) lct.sz[i] = 1;
for(int i = 1; i <= m; i++) {
int vi = read(), vj = read();
ln[i + n] = vi, rn[i + n] = vj;
int x = read();
lct.val[i + n] = x;
lct.pushup(i + n);
if(lct.findr(vi) != lct.findr(vj)) {
if(lct.sum(vi)) cnt--;
if(lct.sum(vj)) cnt--;
lct.link(vi, i + n);
lct.link(i + n, vj);
lct.split(vi, vj);
if(lct.sum(vj)) cnt++;
} else {
int pl = lct.query(vi, vj);
if(lct.val[pl] <= x) {
if(cnt == 0) printf("%d
", st.rbegin()->first);
else puts("-1");
continue;
}
lct.cut(ln[pl], pl);
lct.cut(rn[pl], pl);
lct.link(vi, i + n);
lct.link(i + n, vj);
// lct.split(vi, vj);
st.erase(st.find(mmp(lct.val[pl], pl)));
}
st.insert(mmp(x, i + n));
if(cnt) {
puts("-1");
continue;
}
while(1) {
it = st.rbegin();
int id = it->second;
lct.cut(ln[id], id);
lct.cut(id, rn[id]);
if(lct.sum(ln[id]) || lct.sum(rn[id])) {
lct.link(ln[id], id);
lct.link(id, rn[id]);
break;
}
st.erase(*it);
}
printf("%d
", st.rbegin()->first);
}
return 0;
}
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