Room Allocation
Key Idea: Sweep arrival/departure events in time order, assigning the smallest currently free room number.
Solution
#include <bits/stdc++.h>
using namespace std;
//author: von_Braun
#define ll long long
#define lli long long int
#define pb push_back
#define rep(var, start, num) for(ulli var = start; var <start + num; var++)
#define all(x) x.begin(), x.end()
#define ulli unsigned long long int
#define ull unsigned long long
bool sortbysec(const pair<ll,ll> &a,const pair<ll,ll> &b) { return (a.second < b.second); }
void solve() {
set<tuple<int,int,int>> S;
int n;
cin>>n;
int a,b;
rep(i,0,n) {
cin>>a>>b;
S.insert({a,0,i});
S.insert({b,1,i});
}
vector<int> rooms(n,-1);
set<int> arooms;
int mx=0;
rep(i,1,n) {arooms.insert(i);}
while(!S.empty()) {
auto [curtime, isdep, idx] = *(S.begin());
S.erase(S.begin());
if (isdep) {
arooms.insert(rooms[idx]);
} else {
int curroom = *arooms.begin();
rooms[idx]=curroom;
mx=max(mx,curroom);
arooms.erase(arooms.begin());
}
}
cout<<mx<<endl;
rep(i,0,n) {cout<<rooms[i]<<" ";}
}
int main() {
//add quotes incase input output file
//freopen(input.txt,r,stdin);
//freopen(output.txt,w,stdout);
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int tc = 1;
// cin >> tc;
for (int t = 1; t <= tc; t++) {
solve();
}
}