This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#define ll long long
#define f first
#define s second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define epb emplace_back
#define ull unsigned ll
using namespace std;
const int nmax = 200001;
vector <int> lazy(4 * nmax);
vector <pii> t(4 * nmax);
void push(int v){
lazy[2 * v] = max(lazy[2 * v], lazy[v]);
t[v * 2].f = max(t[v * 2].f, lazy[v]);
lazy[2 * v + 1] = max(lazy[2 * v + 1], lazy[v]);
t[v * 2 + 1].f = max(t[2 * v + 1].f, lazy[v]);
}
void upd(int v, int l, int r, int st, int fin, int val){
if(l > fin || r < st)
return;
if(l >= st && r <= fin){
lazy[v] = max(lazy[v], val);
t[v].f = max(t[v].f, val);
return;
}
int m = (l + r) / 2;
push(v);
upd(2 * v, l, m, st, fin, val);
upd(2 * v + 1, m + 1, r, st, fin, val);
t[v] = min(t[2 * v], t[2 * v + 1]);
}
pii get(int v, int l, int r, int st, int fin){
if(l > fin || r < st)
return {1e9, 1e9};
if(l >= st && r <= fin){
return t[v];
}
push(v);
int m= (l + r) / 2;
return min(get(2 * v, l, m, st, fin), get(2 * v + 1, m + 1, r, st, fin));
}
int main(){
ios_base::sync_with_stdio(false); cin.tie(0);
int n; cin >> n;
int q; cin >> q;
int l[n], r[n];
fill(l, l + n, -1);
fill(r, r + n, n);
for(int i = 1; i <= q; i++){
int L, R, v; cin >> L >> R >> v;
l[v] = max(l[v], L);
r[v] = min(r[v], R);
upd(1, 0, n - 1, L, R, v);
}
bool good = true;
set <int> unused;
for(int i = 0; i < n; ++i){
if(l[i] > r[i]) good = false;
if(l[i] < 0) unused.insert(i);
}
vector <bool> used(n);
vector <int> ans(n);
for(int i= 0; i < n; i++){
pii x = get(1, 0, n - 1, i, i);
// cout << x.f << ' ';
if(used[x.f] || l[x.f] > i || r[x.f] < i){
auto it = unused.upper_bound(x.f);
if(it == unused.end()) good = false;
else ans[i] = *it, unused.erase(it);
}
else used[x.f] = true, ans[i] = x.f;
}
if(!good){
for(int i = 0; i < n; i++)
cout << -1 << ' ';
cout << "\n";
}
else{
for(int i = 0; i < n; i++){
cout << ans[i] << ' ';
}
}
}
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