oj.uz

답안 #988873

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
988873 2024-05-26T14:12:57 Z arpitpandey992 운세 보기 2 (JOI14_fortune_telling2) C++14
4 / 100
 3000 ms 4052 KB 
const long long M = 1e9 + 7;
const int INF = 2147483647;
const long long INFLL = 9223372036854775807ll;
#pragma region Template Start
#include <algorithm>
#include <chrono>
#include <climits>
#include <cmath>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// template <typename T>
// using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// template <typename T>
// using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using tiii = tuple<int, int, int>;
using tlll = tuple<ll, ll, ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vpii = vector<pii>;
using vpll = vector<pll>;
#define endl '\n'
#define nl cout << '\n'
#define pb push_back
#define pob pop_back
#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second
#define FIX(number, digits) fixed << setprecision(digits) << number  // use in cout
#define fok(i, k, n) for (ll i = k; i < n; i++)
#define Fok(i, k, n) for (ll i = n; i >= k; i--)
#define fo(i, n) for (ll i = 0; i < n; i++)
#define Fo(i, n) for (ll i = n; i >= 0; i--)
#define CHK(s, k) (s.find(k) != s.end())
#define all(v) v.begin(), v.end()
#define allg(v) v.rbegin(), v.rend()
#define Sort(v) sort(all(v))
#define Sortg(v) sort(allg(v))
#define sz(v) (static_cast<ll>(v.size()))
#define bs(v, val) binary_search(all(v), val)
#define lb(v, val) lower_bound(all(v), val)
#define ub(v, val) upper_bound(all(v), val)
#define setbits(x) __builtin_popcount(x)
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now()
#define measure()                                              \
    auto end_time = std::chrono::high_resolution_clock::now(); \
    cerr << (end_time - start_time) / std::chrono::milliseconds(1) << "ms" << endl

#define fastio                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL)
#define fileio                        \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout)

#pragma endregion Template End

class RangeCountHelper {
   public:
    RangeCountHelper(int n) {
        this->n = n;
        this->st.resize(4 * n, 0);
    }

    int getCount(int l, int r) {
        return this->_query(l, r, 0, n - 1, 0);
    }

    void switchOn(int index) {
        this->_update(index, 0, n - 1, 0);
    }

   private:
    vector<int> st;
    int n;

    int _query(int l, int r, int sl, int sr, int idx) {
        if (l > sr || r < sl || sl > sr)
            return 0;
        if (sl >= l && sr <= r)
            return st[idx];
        int mid = (sl + sr) / 2;
        return _query(l, r, sl, mid, idx * 2 + 1) + _query(l, r, mid + 1, sr, idx * 2 + 2);
    }

    void _update(int i, int sl, int sr, int idx) {
        if (i > sr || i < sl)
            return;
        if (sl == sr) {
            st[idx] = 1;
            return;
        }
        int mid = (sl + sr) / 2;
        if (i <= mid)
            _update(i, sl, mid, idx * 2 + 1);
        else
            _update(i, mid + 1, sr, idx * 2 + 2);
        st[idx] = st[idx * 2 + 1] + st[idx * 2 + 2];
    }
};

class LargestIndexHelper {
   public:
    LargestIndexHelper(vector<int> &a) {
        this->n = a.size();
        this->a = a;
        this->st.resize(4 * n, 0);
        this->_build(0, n - 1, 0);
    }

    int getLargestIndex(int greaterThanEqualTo) {
        return _query(greaterThanEqualTo, 0, n - 1, 0);
    }

    void invalidate(int index) {
        this->_remove(index, 0, n - 1, 0);
    }

   private:
    vector<int> st;
    vector<int> a;
    int n;

    int _query(int k, int sl, int sr, int idx) {
        if (st[idx] == -1)
            return -1;
        if (sl == sr) {
            return a[sl] >= k ? sl : -1;
        }
        if (st[idx] != -1 && a[st[idx]] >= k)
            return st[idx];
        int mid = (sl + sr) / 2;
        if (st[idx * 2 + 2] != -1) {  // this being -1 means that the whole subarray has been invalidated, so no point
            int right = _query(k, mid + 1, sr, idx * 2 + 2);
            if (right != -1)
                return right;
        }
        if (st[idx * 2 + 1] != -1)
            return _query(k, sl, mid, idx * 2 + 1);
        return -1;
    }

    void _remove(int i, int sl, int sr, int idx) {
        if (i > sr || i < sl)
            return;
        if (sl == sr) {
            a[i] = -1;     // this will no longer be considered during query
            st[idx] = -1;  // for optimization?
            return;
        }
        int mid = (sl + sr) / 2;
        if (i <= mid)
            _remove(i, sl, mid, idx * 2 + 1);
        else
            _remove(i, mid + 1, sr, idx * 2 + 2);
        st[idx] = max(st[idx * 2 + 1], st[idx * 2 + 2]);
    }

    int _build(int sl, int sr, int idx) {
        if (sl == sr) {
            return st[idx] = sl;
        }
        int mid = (sl + sr) / 2;
        int left = _build(sl, mid, idx * 2 + 1);
        int right = _build(mid + 1, sr, idx * 2 + 2);
        return st[idx] = max(left, right);
    }
};

void solve() {
    int n, k;
    cin >> n >> k;
    vector<pair<int, int>> a(n);
    vector<int> queries(k);
    fo(i, n) {
        cin >> a[i].first >> a[i].second;
    }
    sort(a.begin(), a.end(), [](auto &p1, auto &p2) { return max(p1.ff, p1.ss) > max(p2.ff, p2.ss); });
    fo(i, k) {
        cin >> queries[i];
    }
    RangeCountHelper rangeCountHelper(queries.size());
    LargestIndexHelper largestIndexHelper(queries);
    vector<pair<int, int>> q;
    for (int i = 0; i < k; i++) {
        q.push_back({queries[i], i});
    }
    sort(q.begin(), q.end());
    ll ans = 0;
    for (auto &[ai, bi] : a) {
        while (q.size() && q.back().ff >= max(ai, bi)) {
            auto [currentMaxQuery, currentMaxQueryIndex] = q.back();
            q.pop_back();
            rangeCountHelper.switchOn(currentMaxQueryIndex);
            largestIndexHelper.invalidate(currentMaxQueryIndex);
        }
        int largestIndex = largestIndexHelper.getLargestIndex(min(ai, bi));
        int rotationCount = rangeCountHelper.getCount(largestIndex + 1, k - 1);
        bool isRotated = rotationCount % 2 == 1;
        if (ai < bi) {
            if (largestIndex != -1)
                isRotated = !isRotated;
        }
        ans += isRotated ? (ll)bi : (ll)ai;
    }
    cout << ans << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll tes = 1;
    // cin >> tes;
    for (ll t = 1; t <= tes; t++) {
        // cout << "Case #" << t << ": ";
        solve();
    }
}
/*
    What to do?
    1. for any query q[i] >= a[i].ss && q[i] < a[i].ff (assuming first > second)
        - This will cause rotation to revert back to initial state (above assumption)
        - After last q[i] >= ... && q[i] < ..., all q[i] >= a[i].ff will rotate once
    2. find the last query index whose value is lesser than a[i].ff but >= a[i].ss
        - max index in array whose value is >= given value [queries]
        - since we are iterating in reducing value of a[i].ff, we can permanently ignore them when calculating maxIndex
        - basically largest index with value >= given query
    3. rotationCount = number of q[i] where i > above index && q[i] >= a[i].ff
        - willRotate = rotationCount%2 == 1
        - if a[i].ff < a[i].ss:
            - rotation will still count in the same way, just that we assume we start from bigger value as first
            - if no query between a[i].ff and a[i].ss, then rotation will count from smaller value (edge case).
    4. when iterating in reverse manner of a[i].ff, the above count will only increase
        - simple range sum query in binary array
        - initially, all zero
        - while iterating on a, make imaginaryArray[j] = 1 where q[j] >= a[i].ff
*/

Compilation message

fortune_telling2.cpp:4: warning: ignoring '#pragma region Template' [-Wunknown-pragmas]
    4 | #pragma region Template Start
      | 
fortune_telling2.cpp:87: warning: ignoring '#pragma endregion Template' [-Wunknown-pragmas]
   87 | #pragma endregion Template End
      | 
fortune_telling2.cpp: In function 'void solve()':
fortune_telling2.cpp:221:16: warning: structured bindings only available with '-std=c++17' or '-std=gnu++17'
  221 |     for (auto &[ai, bi] : a) {
      |                ^
fortune_telling2.cpp:223:18: warning: structured bindings only available with '-std=c++17' or '-std=gnu++17'
  223 |             auto [currentMaxQuery, currentMaxQueryIndex] = q.back();
      |                  ^

Subtask #1
4.0 / 4.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 348 KB Output is correct
2 Correct 1 ms 480 KB Output is correct
3 Correct 1 ms 348 KB Output is correct
4 Correct 1 ms 504 KB Output is correct
5 Correct 1 ms 348 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 3 ms 552 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 1 ms 472 KB Output is correct
10 Correct 1 ms 344 KB Output is correct
11 Correct 2 ms 348 KB Output is correct
12 Correct 1 ms 348 KB Output is correct
13 Correct 2 ms 348 KB Output is correct

Subtask #2
0 / 31.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 348 KB Output is correct
2 Correct 1 ms 480 KB Output is correct
3 Correct 1 ms 348 KB Output is correct
4 Correct 1 ms 504 KB Output is correct
5 Correct 1 ms 348 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 3 ms 552 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 1 ms 472 KB Output is correct
10 Correct 1 ms 344 KB Output is correct
11 Correct 2 ms 348 KB Output is correct
12 Correct 1 ms 348 KB Output is correct
13 Correct 2 ms 348 KB Output is correct
14 Correct 7 ms 1372 KB Output is correct
15 Correct 13 ms 2264 KB Output is correct
16 Correct 20 ms 2796 KB Output is correct
17 Correct 27 ms 3808 KB Output is correct
18 Correct 28 ms 3808 KB Output is correct
19 Correct 22 ms 3808 KB Output is correct
20 Correct 38 ms 3772 KB Output is correct
21 Correct 18 ms 3808 KB Output is correct
22 Correct 16 ms 3552 KB Output is correct
23 Correct 20 ms 3556 KB Output is correct
24 Correct 21 ms 3548 KB Output is correct
25 Correct 14 ms 3544 KB Output is correct
26 Correct 1373 ms 3536 KB Output is correct
27 Correct 1992 ms 3808 KB Output is correct
28 Correct 855 ms 4052 KB Output is correct
29 Execution timed out 3014 ms 3804 KB Time limit exceeded
30 Halted 0 ms 0 KB -

Subtask #3
0 / 65.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 348 KB Output is correct
2 Correct 1 ms 480 KB Output is correct
3 Correct 1 ms 348 KB Output is correct
4 Correct 1 ms 504 KB Output is correct
5 Correct 1 ms 348 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 3 ms 552 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 1 ms 472 KB Output is correct
10 Correct 1 ms 344 KB Output is correct
11 Correct 2 ms 348 KB Output is correct
12 Correct 1 ms 348 KB Output is correct
13 Correct 2 ms 348 KB Output is correct
14 Correct 7 ms 1372 KB Output is correct
15 Correct 13 ms 2264 KB Output is correct
16 Correct 20 ms 2796 KB Output is correct
17 Correct 27 ms 3808 KB Output is correct
18 Correct 28 ms 3808 KB Output is correct
19 Correct 22 ms 3808 KB Output is correct
20 Correct 38 ms 3772 KB Output is correct
21 Correct 18 ms 3808 KB Output is correct
22 Correct 16 ms 3552 KB Output is correct
23 Correct 20 ms 3556 KB Output is correct
24 Correct 21 ms 3548 KB Output is correct
25 Correct 14 ms 3544 KB Output is correct
26 Correct 1373 ms 3536 KB Output is correct
27 Correct 1992 ms 3808 KB Output is correct
28 Correct 855 ms 4052 KB Output is correct
29 Execution timed out 3014 ms 3804 KB Time limit exceeded
30 Halted 0 ms 0 KB -