Submission #59186

#	Time	Username	Problem	Language	Result	Execution time	Memory
59186		Benq	Teams (CEOI11_tea)	C++14	100 / 100	878 ms	106996 KiB

tea

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 1000001;

int n, pre[MX], res[MX];
pi v[MX];
vi posi; 

struct {
    int mx, par[MX], lft[MX];
    
    int get(int x) {
        if (par[x] != x) par[x] = get(par[x]);
        return par[x];
    }
    
    void ins(int lst) {
        mx = lst;
        par[lst] = lft[lst] = lst;
        while (lft[lst] > 0 && res[lft[lst]-1] <= res[lst]) {
            par[lft[lst]-1] = lst;
            lft[lst] = lft[lft[lst]-1];
        }
    }
    
    pi query(int x) {
        if (x > mx) return {-MOD,-MOD};
        x = get(x);
        return {res[x],x};
    }
} D;

int get(int mid) {
    D.mx = -1;
    int lst = -1;
    for (int i: posi) { // instead of processing in increasing i, increasing i-v[i].f?
        while (lst < i-v[i].f) D.ins(++lst);
        pi x = D.query(max(i-mid,0)); if (x.f != -MOD) x.f ++;
        pre[i] = x.s; res[i] = x.f;
    }
    return res[n];
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    FOR(i,1,n+1) {
        cin >> v[i].f;
        v[i].s = i;
    }
    sort(v+1,v+n+1);
    FOR(i,1,n+1) posi.pb(i);
    sort(all(posi),[](int a, int b) { return a-v[a].f < b-v[b].f; });
    int x = get(n);
    int lo = 1, hi = n;
    while (lo < hi) {
        int mid = (lo+hi)/2;
        if (get(mid) == x) hi = mid;
        else lo = mid+1;
    }
    get(lo);
    int cur = n;
    vector<vi> ans;
    while (cur) {
        int CUR = pre[cur];
        vi ret; FOR(i,CUR+1,cur+1) ret.pb(v[i].s); ans.pb(ret);
        cur = CUR;
    }
    cout << sz(ans) << "\n";
    for (auto a: ans) {
        cout << sz(a) << " ";
        for (int i: a) cout << i << " ";
        cout << "\n";
    }
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10