제출 #1268851

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1268851		MunkhErdene	Art Exhibition (JOI18_art)	C++17	0 / 100	14 ms	31808 KiB

art

#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long
#define lll __int128
#define ll long long
const ll mod = 1e9 + 7;
const ll mod1 = 998244353;
const ll naim = 1e9;
const ll max_bit = 60;
const ull tom = ULLONG_MAX;
const ll MAXN = 100005;
const ll LOG = 20;
const ll NAIM = 1e18;
const ll N = 2e6 + 5;
int main() {
    #define pb push_back
    #define ff first
    #define ss second
    #define _ << " " <<
    #define yes cout<<"YES\n"
    #define no cout<<"NO\n"
    #define all(x) x.begin(),x.end()
    #define rall(x) x.rbegin(),x.rend()
    #define BlueCrowner ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    #define FOR(i, a, b) for (ll i = (a); i < (b); i++)
    #define FORD(i, a, b) for (ll i = (a); i >= (b); i--)
    // ---------- GCD ----------
    auto gcd = [&](ll a, ll b) {
        while (b) {
            a %= b;
            swap(a, b);
        }
        return a;
    };
    // ---------- LCM ----------
    auto lcm = [&](ll a, ll b) {
        return a / gcd(a, b) * b;
    };
    // ---------- Modular Exponentiation ----------
    function<ll(ll, ll, ll)> modpow = [&](ll a, ll b, ll m) {
        ll c = 1;
        a %= m;
        while (b > 0) {
            if (b & 1) c = c * a % m;
            a = a * a % m;
            b >>= 1;
        }
        return c;
    };
    // ---------- Modular Inverse (Fermat’s Little Theorem) ----------
    function<ll(ll, ll)> modinv = [&](ll a, ll m) {
        return modpow(a, m - 2, m);
    };
    // ---------- Factorials and Inverse Factorials ----------
    vector<ll> fact(N), invfact(N);
    auto pre_fact = [&](ll n = N-1, ll m = mod) {
        fact[0] = 1;
        for (ll i = 1; i <= n; i++) fact[i] = fact[i-1] * i % m;
        invfact[n] = modinv(fact[n], m);
        for (ll i = n; i > 0; i--) invfact[i-1] = invfact[i] * i % m;
    };
    // ---------- nCr ----------
    auto nCr = [&](ll n, ll r, ll m = mod) {
        if (r < 0 || r > n) return 0LL;
        return fact[n] * invfact[r] % m * invfact[n-r] % m;
    };
    // ---------- Sieve of Eratosthenes ----------
    vector<ll> primes;
    vector<bool> is_prime(N);
    auto sieve = [&](ll n = N-1) {
        fill(is_prime.begin(), is_prime.begin() + n + 1, true);
        is_prime[0] = is_prime[1] = false;
        for (ll i = 2; i * i <= n; i++) {
            if (is_prime[i]) {
                for (ll j = i * i; j <= n; j += i)
                    is_prime[j] = false;
            }
        }
        for (ll i = 2; i <= n; i++)
            if (is_prime[i]) primes.pb(i);
    };
    function<void()> solve = [&]() {
        ll n; cin >> n;
        map<ll, ll> mp;
        FOR(i, 0, n){
            ll x, y; cin >> x >> y;
            mp[x] += y;
        }
        vector<pair<ll, ll>> a;
        for(auto [x, y] : mp) a.pb({x, y});
        ll m = a.size();
        ll s = 0;
        FOR(i, 0, m) s += a[i].ss;
        s -= a[m - 1].ff - a[0].ff;
        vector<ll> dp(m, 0);
        ll ans = s;
        FOR(i, 1, m){
            dp[i] = dp[i - 1] - a[i - 1].ss + a[i].ff - a[i - 1].ff;
        }
        
        vector<ll> dp1(m, 0);
        FORD(i, m - 2, 0){
            dp1[i] = dp1[i + 1] - a[i + 1].ss + a[i + 1].ff - a[i].ff;
        }
        vector<ll> dpmx(m, 0);
        vector<ll> dp1mx(m, 0);
        dpmx[0] = 0;
        dpmx[1] = dp[0];
        FOR(i, 2, m){
            dpmx[i] = max(dpmx[i - 1], dp[i - 1]);
        }
        dp1mx.back() = 0;
        dp1mx[m - 2] = dp1.back();
        FORD(i, m - 3, 0){
            dp1mx[i] = max(dp1mx[i + 1], dp1[i + 1]);
        }
        /*FOR(i, 0, m){
            cout << a[i].ff _ a[i].ss << '\n';
        }
        cout << '\n';
        FOR(i, 0, m){
            cout << dp[i] <<' ';
        }
        cout << '\n';
        FOR(i, 0, m){
            cout << dpmx[i] << ' ';
        }
        cout << '\n';
        FOR(i, 0, m){
            cout << dp1[i] << ' ';
        }
        cout << '\n';
        FOR(i, 0, m){
            cout << dp1mx[i] << ' ';
        }
        cout << '\n';*/
        ll res = ans;
        FOR(i, 0, m){
            
            res = max(res, ans + dpmx[i] + dp1mx[i]);
        }
        cout << res << '\n';
    };
    BlueCrowner;
    int t = 1; //cin >> t;
    while (t--) {
        solve();
    }
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 20
• Subtask 3: 0 / 20
• Subtask 4: 0 / 50