Submission #1250725

#	Time	Username	Problem	Language	Result	Execution time	Memory
1250725		thegamercoder19	3개의 봉우리 (IOI25_triples)	C++20	51 / 100	2093 ms	100136 KiB

triples

#include <iostream>
#include <vector>
#include <numeric>
#include <map>
#include <set>
#include <tuple>
#include <algorithm>

// Placeholder for Part II, required by the grader to compile.
std::vector<int> construct_range(int M, int K) {
    return {};
}

// Correctly counts mythical triples without overcounting.
long long count_triples(std::vector<int> H) {
    int N = H.size();
    if (N < 3) {
        return 0;
    }
    
    // Use a set of tuples to store unique mythical triples found.
    // This elegantly solves the overcounting problem.
    std::set<std::tuple<int, int, int>> found_triples;

    // We check all 6 permutations. For each valid triple (i, j, k) found,
    // we insert it into the set. The set ensures uniqueness.
    
    // Iterate through the middle element `j` to structure the search
    for (int j = 1; j < N - 1; ++j) {
        // --- Find potential i's (i < j) ---
        // Case where H[j] = j - i  =>  i = j - H[j]
        if (H[j] > 0 && j - H[j] >= 0) {
            int i = j - H[j];
            if (i < j) {
                // Now we have (i, j) and need to find k > j
                
                // Case 3: H[i]=b, H[j]=a, H[k]=a+b
                int k = j + H[i];
                if (k > j && k < N && H[k] == H[i] + H[j]) {
                    found_triples.insert({i, j, k});
                }
                
                // Case 5: H[i]=a+b, H[j]=a, H[k]=b
                k = i + H[i];
                if (k > j && k < N && H[i] > H[j] && H[k] == H[i] - H[j]) {
                    found_triples.insert({i, j, k});
                }
            }
        }
    }

    // --- Forward pass for cases anchored at i ---
    for (int i = 0; i < N; ++i) {
        // Case where H[i] = j - i  =>  j = i + H[i]
        if (H[i] > 0 && i + H[i] < N) {
            int j = i + H[i];
            if (j > i) {
                // Now we have (i, j) and need to find k > j
                
                // Case 1: H[i]=a, H[j]=b, H[k]=a+b
                if (H[j] > 0) {
                    int k = j + H[j];
                    if (k > j && k < N && H[k] == H[i] + H[j]) {
                        found_triples.insert({i, j, k});
                    }
                }

                // Case 2: H[i]=a, H[j]=a+b, H[k]=b
                int k = i + H[j];
                if (k > j && k < N && H[j] > H[i] && H[k] == H[j] - H[i]) {
                    found_triples.insert({i, j, k});
                }
            }
        }
    }
    
    // --- Backward pass for cases anchored at k ---
    for (int k = N-1; k >= 0; --k) {
        // Case where H[k] = k - j => j = k - H[k]
        if(H[k] > 0 && k - H[k] >= 0) {
            int j = k - H[k];
            if (j < k) {
                // Now we have (j, k) and need to find i < j
                
                // Case 6: H[i]=a+b, H[j]=b, H[k]=a
                int i = k - H[i]; // This is not right, H[i] is unknown.
                                 // Let's re-derive: k-i = H[i], so k = i+H[i]
                // This case is difficult to anchor at k.
            }
        }
    }
    
    // A more robust map-based approach for the remaining complex cases
    // to ensure all are found without overcounting.
    std::map<int, std::vector<int>> i_plus_Hi;
    std::map<int, std::vector<int>> k_minus_Hk;
    for(int i = 0; i < N; ++i) i_plus_Hi[i+H[i]].push_back(i);
    for(int k = 0; k < N; ++k) k_minus_Hk[k-H[k]].push_back(k);

    // Case 4: H[i]=b, H[j]=a+b, H[k]=a.  =>  k-i=H[j], j-i=H[k], k-j=H[i]
    // From this: j = i + H[k] and k = j + H[i] => k-H[i] = i+H[k]
    for(auto const& [val, i_list] : i_plus_Hi) {
        if(k_minus_Hk.count(val)) {
            for(int i : i_list) {
                for(int k : k_minus_Hk[val]) {
                    if (i < k) {
                        int j = i + H[k];
                        if (j > i && j < k && H[j] == H[i] + H[k]) {
                           found_triples.insert({i, j, k});
                        }
                    }
                }
            }
        }
    }
    
    // Case 6: H[i]=a+b, H[j]=b, H[k]=a => k-i=H[i], j-i=H[k], k-j=H[j]
    // From this: k=i+H[i] and j=i+H[k]
    // And k-H[j]=j => k=j+H[j]
    // So we need i+H[i] = j+H[j] = k
    for(auto const& [k, i_list] : i_plus_Hi) {
        if (k >= 0 && k < N) { // k must be a valid index
            for(int i : i_list) {
                // Now find a `j` such that j+H[j] = k and i < j < k
                if(i_plus_Hi.count(k)) {
                    for(int j : i_plus_Hi[k]) {
                        if (i < j && j < k) { // Ensure order
                           if (H[i] > H[j] && H[k] == H[i] - H[j]) {
                               found_triples.insert({i, j, k});
                           }
                        }
                    }
                }
            }
        }
    }

    return found_triples.size();
}

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