제출 #73610

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
73610		Benq	Arranging Tickets (JOI17_arranging_tickets)	C++14	85 / 100	2369 ms	13656 KiB

arranging_tickets

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

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

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
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 = 200001;

int N,M, fst, lst;
ll pre[MX], PRE[MX];
vpi tri[MX];

bool test(ll mx, int pos, ll numFlips) {
    FOR(i,1,N+1) PRE[i] = 0;
    ll already = 0;
    priority_queue<array<int,3>> p; // end, begin, num
    FOR(i,1,pos+1) {
        for (auto a: tri[i]) if (a.f > pos) p.push({a.f,i,a.s});
        ll need = pre[i]+numFlips-mx;
        if (i != pos) need = (need+1)/2;
        else need = numFlips;
        need = max(need-already,0LL); already += need; 
        if (already > numFlips) return 0;
        while (need) {
            if (!sz(p)) return 0;
            auto a = p.top(); ll num = min(need,(ll)a[2]);
            a[2] -= num; need -= num; PRE[1] ++; PRE[a[1]] -= 2; PRE[a[0]] += 2;
            p.pop(); if (a[2]) p.push(a);
        }
    }
    FOR(i,1,N+1) PRE[i] += PRE[i-1];
    FOR(i,1,N+1) if (PRE[i]+pre[i] > mx) return 0;
    return 1;
}

bool ok(ll mid) {
    return test(mid,fst,pre[fst]-mid) || test(mid,fst,pre[fst]-mid+1)
    || test(mid,lst,pre[lst]-mid) || test(mid,lst,pre[lst]-mid+1);
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M;
    F0R(i,M) {
        int A,B,C; cin >> A >> B >> C;
        if (A > B) swap(A,B);
        pre[A] += C, pre[B] -= C;
        tri[A].pb({B,C});
    }
    FOR(i,1,N+1) {
        pre[i] += pre[i-1];
        if (pre[i] > pre[fst]) fst = i;
        if (pre[i] >= pre[lst]) lst = i;
    }
    
    ll lo = 0, hi = pre[fst];
    while (lo < hi) {
        ll mid = (lo+hi)/2;
        if (ok(mid)) hi = mid;
        else lo = mid+1;
    }
    cout << lo;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/

• Subtask 1: 10 / 10
• Subtask 2: 35 / 35
• Subtask 3: 20 / 20
• Subtask 4: 20 / 20
• Subtask 5: 0 / 15