oj.uz

답안 #99637

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
99637 2019-03-06T02:13:39 Z jasony123123 Bubble Sort 2 (JOI18_bubblesort2) C++11
0 / 100
 27 ms 1792 KB 
#define _CRT_SECURE_NO_WARNINGS
#include "bubblesort2.h"
#include <bits/stdc++.h>
#include <array>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
//using namespace __gnu_pbds;

#define FOR(i,start,end) for(int i=start;i<(int)(end);i++)
#define FORE(i,start,end) for(int i=start;i<=(int)end;i++)
#define RFOR(i,start,end) for(int i = start; i>end; i--)
#define RFORE(i,start,end) for(int i = start; i>=end; i--)
#define all(a) a.begin(), a.end()
#define mt make_tuple
#define mp make_pair
#define v vector
#define sf scanf
#define pf printf
#define dvar(x) cout << #x << " := " << x << "\n"
#define darr(x,n) FOR(i,0,n) cout << #x << "[" << i << "]" << " := " << x[i] << "\n"

typedef long long ll;
typedef long double ld;
typedef pair<int, int > pii;
typedef pair<ll, ll> pll;
//template <class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> void minn(T &a, T b) { a = min(a, b); }
template<class T> void maxx(T &a, T b) { a = max(a, b); }

void io() {
#ifdef LOCAL_PROJECT 
	freopen("input.in", "r", stdin); freopen("output.out", "w", stdout);
#else 

#endif 
	ios_base::sync_with_stdio(false); cin.tie(NULL);
}
/***************************JOI Open 18 - bubblesort***********************************/

const int inf = 1 << 20;
struct LazySegTree {
	int sz;
	v<int> data, lazy;

	void init(int n, int val) {
		data = v<int>(4 * n, val), lazy = v<int>(4 * n, val);
		sz = n;
	}
	void prop(int ind, int L, int R) {
		if (lazy[ind] != 0) {
			data[ind] += lazy[ind];
			if (L != R) {
				lazy[2 * ind] += lazy[ind];
				lazy[2 * ind + 1] += lazy[ind];
			}
			lazy[ind] = 0;
		}
	}
	void combine(int ind) {
		data[ind] = max(data[2 * ind], data[2 * ind + 1]);
	}

	int query() {
		return query(0, sz - 1, 1, 0, sz - 1);
	}
	void update(int lo, int hi, int val) {
		update(lo, hi, val, 1, 0, sz - 1);
	}

	void update(int lo, int hi, int val, int ind, int L, int R) {
		prop(ind, L, R);
		if (hi < L || R < lo) return;
		if (lo <= L && R <= hi) {
			lazy[ind] += val;
			prop(ind, L, R);
			return;
		}
		int M = (L + R) / 2;
		update(lo, hi, val, 2 * ind, L, M); update(lo, hi, val, 2 * ind + 1, M + 1, R);
		combine(ind);
	}
	int query(int lo, int hi, int ind, int L, int R) {
		prop(ind, L, R);
		if (lo > R || L > hi) return -inf;
		if (lo <= L && R <= hi) return data[ind];
		int M = (L + R) / 2;
		return max(query(lo, hi, 2 * ind, L, M), query(lo, hi, 2 * ind + 1, M + 1, R));
	}
};

v<int> countScans(v<int> A, v<int> X, v<int> V) {
	int n = A.size(), q = X.size();

	// compress
	v<pii> dat;
	dat.reserve(n + q);
	FOR(i, 0, n) dat.push_back({ A[i], i });
	FOR(i, 0, q) dat.push_back({ V[i], X[i] });
	sort(all(dat));
	dat.resize(unique(all(dat)) - dat.begin());
	FOR(i, 0, n) A[i] = lower_bound(all(dat), mp(A[i], i)) - dat.begin();
	FOR(i, 0, q) V[i] = lower_bound(all(dat), mp(V[i], X[i])) - dat.begin();

	// init
	LazySegTree tree;
	tree.init(dat.size(), -inf);
	FOR(i, 0, n) {
		tree.update(A[i] + 1, tree.sz - 1, -1);
		tree.update(A[i], A[i], inf+i);
	}

	// process q
	v<int> ans(q);
	FOR(i, 0, q) {
		int pos = X[i];
		int _new = V[i];
		int _old = A[pos];

		tree.update(_old, _old, -inf - pos);
		tree.update(_new, _new, inf + pos);
		if (_new < _old)
			tree.update(_new + 1, _old, -1);
		else if (_new > _old)
			tree.update(_old + 1, _new, +1);

		A[pos] = _new;
		ans[i] = tree.query();
	}

	return ans;
}

Subtask #1
0 / 17.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 4 ms 384 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #2
0 / 21.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 4 ms 384 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #3
0 / 22.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 27 ms 1792 KB Output isn't correct
2 Halted 0 ms 0 KB -

Subtask #4
0 / 40.0

# 결과 실행 시간 메모리 Grader output
1 Incorrect 4 ms 384 KB Output isn't correct
2 Halted 0 ms 0 KB -