| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 1264119 | aihoi | Long Mansion (JOI17_long_mansion) | C++20 | 0 ms | 0 KiB |
#include<bits/stdc++.h>
#include<algorithm>
#include<random>
#include<chrono>
#include<cstdlib>
#include<ctime>
#include<numeric>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<queue>
#include<iomanip>
#define int long long
#define ll long long
#define L LLONG_MAX
#define fi first
#define se second
#define pii pair<int,int>
#define sz(a) ((int)a.size())
#define FOR(i,j,k) for(int i=j;i<=k;i++)
#define REP(i,k,j) for(int i=k;i>=j;i--)
#define FORD(i,a) for(auto i:a)
using namespace std;
const int NMAX = 500000 + 5; // sửa tuỳ bài; JOI limit up to 5e5
int n,q;
int c[NMAX];
vector<int> a[NMAX];
vector<int> pos[NMAX]; // pos[type] = sorted list of rooms that chứa key "type"
int lef[NMAX], rig[NMAX];
bool inside(int l, int r, int key){
if (key <= 0 || key >= NMAX) return false;
auto &v = pos[key];
if (v.empty()) return false;
auto it = lower_bound(v.begin(), v.end(), l);
return it!=v.end() && *it <= r;
}
void input(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
if(!(cin>>n)) exit(0);
FOR(i,1,n-1) cin>>c[i];
FOR(i,1,n){
int b; cin>>b;
a[i].resize(b);
FOR(j,0,b-1) {
cin>>a[i][j];
if (a[i][j] >= 0 && a[i][j] < NMAX) pos[a[i][j]].push_back(i);
}
}
// pos[*] are filled in increasing room order already (we pushed i ascending),
// so they are sorted -> no need to sort again.
cin>>q;
}
void solve(){
// compute lef/rig for each starting room i (1..n)
FOR(i,1,n){
lef[i]=rig[i]=i;
bool nxt = true;
while(nxt){
nxt = false;
// try extend left by checking corridor (lef[i]-1) if exists
while (lef[i] > 1 && inside(lef[i], rig[i], c[lef[i]-1]) ){
// when extend left into (lef[i]-1), we can merge with precomputed segment of lef[i]-1
int L = lef[lef[i]-1];
int R = rig[lef[i]-1];
// update segment
lef[i] = L;
if (R > rig[i]) rig[i] = R;
nxt = true;
}
// try extend right by checking corridor rig[i]
// Note: use if (rig[i] < n) to avoid out-of-bound c[n]
while (rig[i] < n && inside(lef[i], rig[i], c[rig[i]])){
rig[i]++; // include room rig[i]
// after including new room rig[i], there could be previous computed info if rig[i] <= i?
// But we only have lef/rig for indices < current i guaranteed; if rig[i] <= i nothing to merge.
// However if we moved rig[i] to some j < i that has been computed, we can merge:
if (rig[i] < i){
// merge with its precomputed block
if (lef[rig[i]] < lef[i]) lef[i] = lef[rig[i]];
if (rig[rig[i]] > rig[i]) rig[i] = rig[rig[i]];
}
nxt = true;
}
// one more attempt: maybe after right expansion we can again extend left via merged info
// the outer while(nxt) takes care of repeating until stable
}
}
// answer queries
while(q--){
int x,y; cin>>x>>y;
if (x<=y){
if (lef[x] <= y && y <= rig[x]) cout<<"YES\n"; else cout<<"NO\n";
}else{
// if x > y, since segments are contiguous, check same condition (we still computed lef[x],rig[x])
if (lef[x] <= y && y <= rig[x]) cout<<"YES\n"; else cout<<"NO\n";
}
}
}
signed main(){
input();
solve();
return 0;
}