제출 #912061

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
912061		hhngriver	Distributing Candies (IOI21_candies)	C++17	0 / 100	120 ms	57584 KiB

candies

#include <bits/stdc++.h>
#include "candies.h"

using namespace std;

#define FILENAME "0-01.in"
#define MAX_N 50
#define MAX_Q 50

const int n_bits = 4;
const long long inf = 1e18;
long long minseg[1 << (n_bits + 1)];
long long maxseg[1 << (n_bits + 1)];
long long lazyadd[1 << (n_bits + 1)];

// a standard lazy propagation segment tree
// here we need to support both min and max
// so it is essentially 2 segtrees combined together
// but we only need 1 copy of lazy add
struct segtree {
	long long last_value = 0;
	long long small = inf;
	long long big = -inf;

	segtree() {
	}

	void update(int node, int change) { // treated as a suffix update
		last_value += change;
		node += (1 << n_bits);
		lazyadd[node] += change;
		while (node > 1) {
			if (node % 2 == 0) {
				lazyadd[node + 1] += change;
			}
			minseg[node / 2] = min(minseg[node] + lazyadd[node],
					minseg[node ^ 1] + lazyadd[node ^ 1]);
			maxseg[node / 2] = max(maxseg[node] + lazyadd[node],
					maxseg[node ^ 1] + lazyadd[node ^ 1]);
			cout << node << " " << (node ^ 1) << endl;
			node = node / 2;
		}
	}

	int solve(int capacity) { // returns the largest index i, such that the range >= c
		int node = 1;
		small = inf;
		big = -inf;
		long long lz = 0;
		while (node < (1 << n_bits)) {
			lz += lazyadd[node];
			node *= 2;
			if (max(big, maxseg[node + 1] + lazyadd[node + 1] + lz)
					- min(small, minseg[node + 1] + lazyadd[node + 1] + lz)
					> capacity) {
				node++;
			} else {
				big = max(big, maxseg[node + 1] + lazyadd[node + 1] + lz);
				small = min(small, minseg[node + 1] + lazyadd[node + 1] + lz);
			}
		}
		if (minseg[node] + lazyadd[node] + lz < last_value) {
			return capacity - (big - last_value);
		} else {
			return last_value - small;
		}
	}
};

vector<pair<int, int>> toggle[(int) 6e5];
// this tells you what you need to toggle on/off as you move across the boxes
// stores a pair indicating the query id and the change in number of candies
vector<int> distribute_candies(vector<int> C, vector<int> L, vector<int> R,
		vector<int> V) {
	int n = C.size();
	int q = L.size();
	segtree s;

	for (int i = 0; i < q; i++) {
		toggle[L[i]].push_back(make_pair(i, V[i]));
		toggle[R[i] + 1].push_back(make_pair(i, -V[i]));
	}

	vector<int> ans;
	ans.resize(n);
	for (int i = 0; i < n; i++) {
		for (pair<int, int> p : toggle[i]) {
			s.update(p.first + 2, p.second); // store values as if the boxes have infinite capacity
		}

		if (maxseg[1] - minseg[1] < C[i]) { // easy case: range is small
			ans[i] = s.last_value - (minseg[1] + lazyadd[1]);
		} else { // we binary search on the segtree
			ans[i] = s.solve(C[i]);
		}
	}
	return ans;
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 8
• Subtask 3: 0 / 27
• Subtask 4: 0 / 29
• Subtask 5: 0 / 33