Submission #59185

#	Time	Username	Problem	Language	Result	Execution time	Memory
59185		Benq	Teams (CEOI11_tea)	C++11	70 / 100	2559 ms	30680 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];
pi v[MX];

template<class T, int SZ> struct Seg {
    T seg[2*SZ], MN = {-MOD,-MOD};
    
    Seg() {
        memset(seg,0,sizeof seg);
    }
    
    T comb(T a, T b) { return max(a,b); } // easily change this to min or max
    
    void upd(int p, T value) {  // set value at position p
        for (seg[p += SZ] = value; p > 1; p >>= 1)
            seg[p>>1] = comb(seg[(p|1)^1],seg[p|1]); // non-commutative operations
    }
    
    void build() {
        F0Rd(i,SZ) seg[i] = comb(seg[2*i],seg[2*i+1]);
    }
    
    T query(int l, int r) {  // sum on interval [l, r]
        T res1 = MN, res2 = MN; r++;
        for (l += SZ, r += SZ; l < r; l >>= 1, r >>= 1) {
            if (l&1) res1 = comb(res1,seg[l++]);
            if (r&1) res2 = comb(seg[--r],res2);
        }
        return comb(res1,res2);
    }
};

Seg<pi,1<<20> S;

int get(int mid) {
    FOR(i,1,n+1) {
        pi x = S.query(max(i-mid,0),i-v[i].f);
        if (x.f != -MOD) x.f ++;
        pre[i] = x.s;
        S.upd(i,{x.f,i});
        if (i == n) return x.f;
    }
}

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);
    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);
    /*cout << lo << "\n";
    FOR(i,1,n+1) cout << pre[i] << " ";
    cout << "\n";*/
    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
*/

Compilation message (stderr)

tea.cpp: In function 'int get(int)':
tea.cpp:85:1: warning: control reaches end of non-void function [-Wreturn-type]
 }
 ^

• 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: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10