제출 #234074

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
234074		mode149256	마상시합 토너먼트 (IOI12_tournament)	C++14	0 / 100	116 ms	11384 KiB

tournament

/*input

*/
#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;

namespace my_template {
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<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}
}
using namespace my_template;

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;

typedef tree <
pi,
null_type,
less<pi>,
rb_tree_tag,
tree_order_statistics_node_update >
ordered_set;

vpi convert(int N, int C, int *S, int *E) {
	ordered_set curr;
	vpi queries;
	for (int i = 0; i < N; ++i)
		curr.insert({i, i});

	for (int i = 0; i < C; ++i)
	{
		auto prad = curr.find_by_order(S[i]);

		pi p;

		p.x = (*prad).x;

		for (int j = 0; j < E[i] - S[i]; ++j)
		{
			auto nxt = next(prad);
			curr.erase(prad);
			prad = nxt;
		}

		p.y = (*prad).y;

		curr.erase(prad);
		curr.insert(p);
		queries.push_back(p);
	}

	return queries;
}


struct max_seg {
	int l, r;
	int did;
	max_seg *left = nullptr;
	max_seg *right = nullptr;

	max_seg(int a, int b, int *K): l(a), r(b) {
		if (l == r) {
			did = K[l];
		} else {
			left = new max_seg(l, (l + r) / 2, K);
			right = new max_seg((l + r) / 2 + 1, r, K);
			did = max(left->did, right->did);
		}
	}

	int get(int a, int b) {
		if (r < a or b < l) return -MOD;
		else if (a <= l and r <= b) return did;
		else
			return max(left->get(a, b), right->get(a, b));
	}
};

struct node {
	int l, r;
	int did;
	int lazy;

	node *left = nullptr;
	node *right = nullptr;

	node(int a, int b): l(a), r(b), did(0), lazy(0) {
		if (l != r) {
			left = new node(l, (l + r) / 2);
			right = new node((l + r) / 2 + 1, r);
		}
	}

	inline void push() {
		if (left) left->lazy += lazy;
		if (right) right->lazy += lazy;

		did += lazy;
		lazy = 0;
	}

	void add(int a, int b, int val) {
		push();
		if (r < a or b < l) return;
		else if (a <= l and r <= b) {
			lazy += val;
			push();
		} else {
			left->add(a, b, val);
			right->add(a, b, val);

			left->push();
			right->push();
			did = max(left->did, right->did);
		}
	}
};

int GetBestPosition(int N, int C, int R, int *K, int *S, int *E) {
	vpi queries = convert(N, C, S, E);

	// for (auto u : queries)
	// 	printf("%d %d\n", u.x, u.y);

	int ats = 0;

	max_seg maxS(0, N - 2, K);

	node med(0, N - 1);
	for (int i = 0; i < C; ++i)
	{
		int s = queries[i].x;
		int e = queries[i].y;

		int d = maxS.get(s, e - 1);

		// printf("s = %d, e = %d, d = %d, R = %d\n", s, e, d, R);
		if (d > R)
			med.add(s, e, -1e6);
		else
			med.add(s, e, 1);

		ats = max(ats, med.did);
	}
	return ats;
}

• Subtask 1: 0 / 17
• Subtask 2: 0 / 32
• Subtask 3: 0 / 51