Submission #1249877

#	Time	Username	Problem	Language	Result	Execution time	Memory
1249877		kduckp	3개의 봉우리 (IOI25_triples)	C++20	14.17 / 100	2095 ms	1972 KiB

triples

#include "triples.h"
#include <vector>
#include <algorithm>
#include <unordered_map>
using namespace std;

long long count_triples(vector<int> H) {
    int n = H.size();
    long long res = 0;

    if (n <= 100) {
        for (int i = 0; i < n; ++i)
            for (int j = i + 1; j < n; ++j)
                for (int k = j + 1; k < n; ++k) {
                    vector<int> h = {H[i], H[j], H[k]};
                    vector<int> d = {j - i, k - i, k - j};
                    sort(h.begin(), h.end());
                    sort(d.begin(), d.end());
                    if (h == d) res++;
                }
        return res;
    }

    bool small_value = true;
    for (int h : H)
        if (h > 10) {
            small_value = false;
            break;
        }

    if (small_value) {
        for (int i = 0; i < n; ++i)
            for (int j = i + 1; j < min(n, i + 11); ++j)
                for (int k = j + 1; k < min(n, i + 22); ++k) {
                    vector<int> h = {H[i], H[j], H[k]};
                    vector<int> d = {j - i, k - i, k - j};
                    sort(h.begin(), h.end());
                    sort(d.begin(), d.end());
                    if (h == d) res++;
                }
        return res;
    }

    if (is_sorted(H.begin(), H.end())) {
        for (int j = 1; j < n - 1; ++j) {
            int target = 2 * H[j];
            int left = 0, right = n - 1;
            while (left < j && right > j) {
                if (H[left] + H[right] == target) {
                    vector<int> h = {H[left], H[j], H[right]};
                    vector<int> d = {j - left, right - left, right - j};
                    sort(h.begin(), h.end());
                    sort(d.begin(), d.end());
                    if (h == d) res++;
                    int curr_left = left, curr_right = right;
                    while (left < j && H[left] == H[curr_left]) left++;
                    while (right > j && H[right] == H[curr_right]) right--;
                } else if (H[left] + H[right] < target) {
                    left++;
                } else {
                    right--;
                }
            }
        }
        return res;
    }

    // --- THÊM XỬ LÝ NÀY ---
    if (n <= 2000) {
        vector<pair<int, int>> arr;
        for (int i = 0; i < n; ++i)
            arr.emplace_back(i, H[i]);

        sort(arr.begin(), arr.end(), [](auto &a, auto &b) {
            return a.second < b.second;
        });

        for (int x = 0; x < n; ++x) {
            int i = arr[x].first;
            for (int y = x + 1; y < n; ++y) {
                int j = arr[y].first;
                if (i >= j) continue;
                vector<int> h = {H[i], H[j]};
                vector<int> d = {j - i};
                for (int k = j + 1; k < n; ++k) {
                    h.push_back(H[k]);
                    d = {j - i, k - i, k - j};
                    vector<int> tmp_h = h;
                    vector<int> tmp_d = d;
                    sort(tmp_h.begin(), tmp_h.end());
                    sort(tmp_d.begin(), tmp_d.end());
                    if (tmp_h == tmp_d) res++;
                    h.pop_back();
                }
            }
        }
        return res;
    }

    // fallback cuối
    unordered_map<int, vector<int>> height_positions;
    for (int i = 0; i < n; i++) {
        height_positions[H[i]].push_back(i);
    }

    for (int i = 0; i < min(n, 1000); ++i) {
        for (int j = i + 1; j < min(n, i + 1000); ++j) {
            for (int k = j + 1; k < min(n, i + 2000); ++k) {
                vector<int> h = {H[i], H[j], H[k]};
                vector<int> d = {j - i, k - i, k - j};
                sort(h.begin(), h.end());
                sort(d.begin(), d.end());
                if (h == d) res++;
            }
        }
    }

    return res;
}

vector<int> construct_range(int M, int K) {
    return {1, 1, 2};
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 6 / 6
• Subtask 3: 0 / 10
• Subtask 4: 0 / 11
• Subtask 5: 0 / 16
• Subtask 6: 0 / 19
• Subtask 7: 0.1667 / 5
• Subtask 8: 0.0025 / 5
• Subtask 9: 0.0001 / 5
• Subtask 10: 0 / 5
• Subtask 11: 0 / 5
• Subtask 12: 0 / 5