답안 #735741

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
735741	2023-05-04T14:17:51 Z	GrindMachine	Kralj (COCI16_kralj)	C++17	140 / 140	579 ms	72696 KB

kralj

// Om Namah Shivaya

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

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
edi
thread: https://codeforces.com/blog/entry/47677?#comment-320681


circular array problems: try to find some cutpoint and transform into linear array problem

find a pos m s.t for any given order, nobody crosses the edge m -> m+1

let cnt(i) = #of elves that are paired up with dwarf x

how to check if given pos m is good?
starting from every pos, we will try our best to create an overflow

following should hold for all pos x:

cnt(x) + cnt(x+1) + ... + cnt(m) <= #of indices in range [x,m] (note: +1 in circular array and range in circular array)
cnt(x) + cnt(x+1) + ... + cnt(m) - (#of indices in range [x,m]) <= 0

(cnt(x)-1) + (cnt(x+1)-1) + ... + (cnt(m)-1) <= 0

replace all cnt(x) with cnt(x)-1,
cnt(x) + cnt(x+1) + ... + cnt(m) <= 0

rewrite in terms of prefix sums
let p(i) = sum[cnt(x)], x <= i (no circular stuff here)

if x > m:
p(n) - p(x) + p(x) = p(n) = 0 (p(n) = 0 because there are n dwarves and n elves)

if x <= m:
p(m) - p(x-1) <= 0

dont care abt first condition cuz p(n) is always 0
we must pick m s.t 2nd condition is satisfied

we can do this by picking m that has min val of p(m), so subtracting it with any other p(x) value would always be <= 0

once m is picked, we can cut the array at point (m,m+1):

now run greedy algo:
consider dwarfs in increasing order of new indices
when considering dwarf, first add all elves that belong to this dwarf to the active set

then, assign the elf with smallest value that has power > power of this dwarf
if such an elf doesnt exist, then assign the elf with lowest power

(idk formal proof, but sounds somewhat convincing)

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> go(n + 5);
    rep1(i, n) cin >> go[i];

    vector<ll> a(n + 5), b(n + 5);
    rep1(i, n) cin >> a[i];
    rep1(i, n) cin >> b[i];

    vector<ll> cnt(n + 5);
    vector<ll> enter[n + 5];

    rep1(i, n) cnt[go[i]]++, enter[go[i]].pb(i);
    rep1(i, n) cnt[i]--;

    vector<ll> pref(n + 5);
    rep1(i, n) pref[i] = pref[i - 1] + cnt[i];

    ll m = min_element(pref.begin() + 1, pref.begin() + n + 1) - pref.begin();

    vector<ll> indices;
    for (int i = m + 1; i <= n; ++i) {
        indices.pb(i);
    }
    rep1(i, m) indices.pb(i);

    set<ll> st;
    ll ans = 0;

    trav(dwarf, indices) {
        trav(elf, enter[dwarf]) {
            st.insert(b[elf]);
        }

        auto it = st.upper_bound(a[dwarf]);
        if (it == st.end()) {
            assert(!st.empty());
            st.erase(st.begin());
        }
        else {
            ans++;
            st.erase(it);
        }
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

Subtask #1

140.0 / 140.0

#	결과	실행 시간	메모리	Grader output
1	Correct	579 ms	59056 KB	Output is correct
2	Correct	345 ms	57344 KB	Output is correct
3	Correct	446 ms	71296 KB	Output is correct
4	Correct	464 ms	72696 KB	Output is correct
5	Correct	312 ms	50996 KB	Output is correct
6	Correct	316 ms	50236 KB	Output is correct
7	Correct	377 ms	54280 KB	Output is correct
8	Correct	322 ms	49436 KB	Output is correct
9	Correct	401 ms	57320 KB	Output is correct
10	Correct	338 ms	54588 KB	Output is correct