oj.uz

답안 #347608

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
347608 2021-01-13T09:14:42 Z BorisBarca Meteors (POI11_met) C++14
0 / 100
 6000 ms 65536 KB 
/*
$$$$$$$\                      $$\           $$$$$$$\
$$  __$$\                     \__|          $$  __$$\
$$ |  $$ | $$$$$$\   $$$$$$\  $$\  $$$$$$$\ $$ |  $$ | $$$$$$\   $$$$$$\   $$$$$$$\ $$$$$$\
$$$$$$$\ |$$  __$$\ $$  __$$\ $$ |$$  _____|$$$$$$$\ | \____$$\ $$  __$$\ $$  _____|\____$$\
$$  __$$\ $$ /  $$ |$$ |  \__|$$ |\$$$$$$\  $$  __$$\  $$$$$$$ |$$ |  \__|$$ /      $$$$$$$ |
$$ |  $$ |$$ |  $$ |$$ |      $$ | \____$$\ $$ |  $$ |$$  __$$ |$$ |      $$ |     $$  __$$ |
$$$$$$$  |\$$$$$$  |$$ |      $$ |$$$$$$$  |$$$$$$$  |\$$$$$$$ |$$ |      \$$$$$$$\\$$$$$$$ |
\_______/  \______/ \__|      \__|\_______/ \_______/  \_______|\__|       \_______|\_______|
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define PB push_back
#define MP make_pair
#define INS insert
#define LB lower_bound
#define UB upper_bound
#define pii pair <int,int>
#define pll pair <long long, long long>
#define X first
#define Y second
#define _ << " " <<
#define sz(x) (int)x.size()
#define all(a) (a).begin(),(a).end()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i > (b); --i)
#define FORA(i, x) for (auto &i : x)
#define REP(i, n) FOR(i, 0, n)
#define BITS(x) __builtin_popcount(x)
#define SQ(a) (a) * (a)
#define TRACE(x) cout << #x " = " << (x) << '\n';
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define umap unordered_map

typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <ll> vll;
typedef vector <pll> vpl;
typedef vector <double> vd;
typedef vector <ld> vld;
typedef vector<string> vs;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
//find_by_order -> kti najmanji
//order_of_key -> koliko ima manjih od x
//((float) t)/CLOCKS_PER_SEC

const int MOD = 1e9 + 7;
const double PI = acos(-1);
const int INF = 1e9 + 10;
const ll INFL = 1e18 + 10;
const int ABC = 30;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dox[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int doy[] = {0, 0, -1, 1, -1, 1, -1, 1};

inline int sum(int a, int b){
	if (a + b < 0)
		return (a + b + MOD) % MOD;
	return (a + b) % MOD;
}

inline void add(int &a, int b){
	a = sum(a, b);
}

inline int mul(ll a, ll b){
	return ((a % MOD) * ((ll)b % MOD)) % MOD;
}

inline int sub(int a, int b){
	return (a - b + MOD) % MOD;
}

inline int fast_pot(ll pot, ll n){
	ll ret = 1;
	while (n){
		if (n & 1LL)
			ret = (ret * pot) % MOD;
		pot = (pot * pot) % MOD;
		n >>= 1LL;
	}
	return ret;
}

inline int divide(int a, int b){
	return mul(a, fast_pot(b, MOD - 2));
}

ll lcm(ll a, ll b){
	return abs(a * b) / __gcd(a, b);
}

inline double ccw(pii A, pii B, pii C){
	return (A.X * B.Y) - (A.Y * B.X) + (B.X * C.Y) - (B.Y * C.X) + (C.X * A.Y) - (C.Y * A.X);
}

inline int CCW(pii A, pii B, pii C){
	double val = ccw(A, B, C);
	double eps = max(max(abs(A.X), abs(A.Y)), max(max(abs(B.X), abs(B.Y)), max(abs(C.X), abs(C.Y)))) / 1e9;
	if (val <= -eps)
		return -1;
	if (val >= eps)
		return 1;
	return 0;
}

void to_upper(string &x){
	REP(i, sz(x))
		x[i] = toupper(x[i]);
}

void to_lower(string &x){
	REP(i, sz(x))
		x[i] = tolower(x[i]);
}

string its(ll x){
	if (x == 0)
		return "0";
	string ret = "";
	while (x > 0){
		ret += (x % 10) + '0';
		x /= 10;
	}
	reverse(all(ret));
	return ret;
}

ll sti(string s){
	ll ret = 0;
	REP(i, sz(s)){
		ret *= 10;
		ret += (s[i] - '0');
	}
	return ret;
}

const int N = 3e5 + 10;
const int LOG = 19;
const int OFF = 1 << LOG;

struct tournament{
	ll t[2 * OFF], lazy[2 * OFF], off;

	void init(int n){
		for (off = 1; off < n; off *= 2);
		memset(t, 0, sizeof t);
		memset(lazy, 0, sizeof lazy);
	}

	void prop(int node){
		t[node * 2] += lazy[node];
		t[node * 2 + 1] += lazy[node];
		lazy[node * 2] += lazy[node];
		lazy[node * 2 + 1] += lazy[node];

		lazy[node] = 0;
	}

	void update(int node, int a, int b, int lo, int hi, ll x){
		if (a > hi || b < lo)
			return;
		if (a >= lo && b <= hi){
			t[node] += x;
			lazy[node] += x;
			return; 
		}
		prop(node);
		int mid = (a + b) / 2;
		update(node * 2, a, mid, lo, hi, x);
		update(node * 2 + 1, mid + 1, b, lo, hi, x);
	}

	ll query(int node, int a, int b, int lo, int hi){
		if (a > hi || b < lo)
			return 0;
		if (a >= lo && b <= hi)
			return t[node];
		prop(node);
		int mid = (a + b) / 2;
		return query(node * 2, a, mid, lo, hi) + query(node * 2 + 1, mid + 1, b, lo, hi); 
	}
} T;

struct query{
	int l, r;
	ll x;
};

ll n, m, o[N], p[N], sol[N];
vector <query> Q;
vi group[N];

void solve(int lo, int hi, vi v){
	if (lo == hi){
		FOR(i, lo, hi + 1){
			if (Q[i].l <= Q[i].r)
				T.update(1, 0, T.off - 1, Q[i].l, Q[i].r, Q[i].x);
			else {
				T.update(1, 0, T.off - 1, Q[i].l, m - 1, Q[i].x);
				T.update(1, 0, T.off - 1, 0, Q[i].r, Q[i].x);
			}
		}

		FORA(x, v){
			ll zbr = 0;
			FORA(el, group[x])
				zbr += T.query(1, 0, T.off - 1, el, el);
			if (zbr >= p[x])
				sol[x] = min(sol[x], (ll)lo);
		}

		FOR(i, lo, hi + 1){
			if (Q[i].l <= Q[i].r)
				T.update(1, 0, T.off - 1, Q[i].l, Q[i].r, -Q[i].x);
			else {
				T.update(1, 0, T.off - 1, Q[i].l, m - 1, -Q[i].x);
				T.update(1, 0, T.off - 1, 0, Q[i].r, -Q[i].x);
			}
		}
		return;
	}

	int mid = (lo + hi) / 2;
	FOR(i, lo, mid + 1){
		if (Q[i].l <= Q[i].r)
			T.update(1, 0, T.off - 1, Q[i].l, Q[i].r, Q[i].x);
		else {
			T.update(1, 0, T.off - 1, Q[i].l, m - 1, Q[i].x);
			T.update(1, 0, T.off - 1, 0, Q[i].r, Q[i].x);
		}
	}
	
	vi l, r;
	FORA(x, v){
		ll zbr = 0;
			FORA(el, group[x])
				zbr += T.query(1, 0, T.off - 1, el, el);
		cout << lo _ mid _ x _ zbr << '\n';
		if (zbr >= p[o[x]]){
			sol[x] = min(sol[x], (ll)mid);
			l.PB(x);
		}
		else
			r.PB(x);
	}

	v.clear();
	solve(mid + 1, hi, r);
	FOR(i, lo, mid + 1){
		if (Q[i].l <= Q[i].r)
			T.update(1, 0, T.off - 1, Q[i].l, Q[i].r, -Q[i].x);
		else {
			T.update(1, 0, T.off - 1, Q[i].l, m - 1, -Q[i].x);
			T.update(1, 0, T.off - 1, 0, Q[i].r, -Q[i].x);
		}
	}
	solve(lo, mid, l);
}

int main () {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> n >> m;
	REP(i, m)
		sol[i] = INF;
	T.init(m);
	REP(i, m){
		cin >> o[i];
		o[i]--;
		group[o[i]].PB(i);
	}

	REP(i, n)
		cin >> p[i];

	int q; cin >> q;
	query ev;
	REP(i, q){
		cin >> ev.l >> ev.r >> ev.x;
		ev.l--;
		ev.r--;
		//cout << ev.l _ ev.r << '\n';
		Q.PB(ev);
	}

	vi idx;
	REP(i, n)
		idx.PB(i);

	solve(0, q - 1, idx);
	REP(i, n){
		if (sol[i] == INF)
			cout << "NIE\n";
		else
			cout << sol[i] + 1 << '\n'; 
	}

		
	return 0;
}

Subtask #1
0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 19 ms 23916 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #2
0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 21 ms 24044 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #3
0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 594 ms 28772 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #4
0 / 12.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 693 ms 31464 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #5
0 / 13.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 400 ms 28772 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #6
0 / 13.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 679 ms 27104 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #7
0 / 13.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 6037 ms 65536 KB Time limit exceeded
2 Halted 0 ms 0 KB -

Subtask #8
0 / 13.0

# 결과 실행 시간 메모리 Grader output
1 Execution timed out 6009 ms 65536 KB Time limit exceeded
2 Halted 0 ms 0 KB -