結果

問題 No.2330 Eat Slime
ユーザー qwewe
提出日時 2025-05-14 12:59:52
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,515 ms / 4,000 ms
コード長 10,310 bytes
コンパイル時間 2,683 ms
コンパイル使用メモリ 111,312 KB
実行使用メモリ 63,556 KB
最終ジャッジ日時 2025-05-14 13:02:20
合計ジャッジ時間 35,539 ms
ジャッジサーバーID
(参考情報)
judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 30
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ソースコード

diff #

#include <iostream>
#include <vector>
#include <complex>
#include <cmath>
#include <numeric>
#include <algorithm>
#include <vector> // Ensure vector is included

// Use long double for potentially better precision in FFT calculations
using namespace std;
typedef long long ll;

// Check compiler support for long double complex types and math functions
#if defined(__GNUC__) || defined(__clang__)
    // Use long double on GCC/Clang environments where it typically offers higher precision
    typedef complex<long double> cld;
    const long double PI_LD = acosl(-1.0L);
    // Macros for math functions to use long double versions
    #define ROUND_FUNC roundl
    #define COS_FUNC cosl
    #define SIN_FUNC sinl
#else
    // Fallback to standard double if long double specifics are unknown or it offers no advantage
    typedef complex<double> cld;
    const double PI_LD = acos(-1.0);
    #define ROUND_FUNC round
    #define COS_FUNC cos
    #define SIN_FUNC sin
#endif


/**
 * @brief Iterative Fast Fourier Transform implementation.
 * 
 * @param a The vector of complex numbers to transform. It is modified in-place.
 * @param invert If true, performs inverse FFT. Otherwise, performs forward FFT.
 * 
 * This implementation uses the Cooley-Tukey algorithm with bit-reversal permutation.
 * It is adapted from standard competitive programming resources.
 */
void fft_iterative(vector<cld>& a, bool invert) {
    int n = a.size();
    if (n == 1) return; // Base case: FFT of size 1 is identity

    // Bit-reversal permutation: reorders elements according to bit-reversed indices
    for (int i = 1, j = 0; i < n; i++) {
        int bit = n >> 1;
        // Find the next index j by reversing the bits of i
        for (; j & bit; bit >>= 1)
            j ^= bit;
        j ^= bit;

        if (i < j) // Swap elements if i < j to avoid double swaps
            swap(a[i], a[j]);
    }

    // Cooley-Tukey algorithm: iteratively combine smaller FFT results
    for (int len = 2; len <= n; len <<= 1) {
        // Angle for the current stage's roots of unity
        long double ang = 2 * PI_LD / len * (invert ? -1 : 1);
        cld wlen(COS_FUNC(ang), SIN_FUNC(ang)); // Principal len-th root of unity
        for (int i = 0; i < n; i += len) { // Iterate through blocks of size len
            cld w(1); // Start with w = 1 (the 0-th power root of unity)
            for (int j = 0; j < len / 2; j++) { // Combine pairs of elements
                cld u = a[i + j]; // Element from first half
                cld v = a[i + j + len / 2] * w; // Element from second half multiplied by root of unity
                a[i + j] = u + v; // Butterfly operation: sum
                a[i + j + len / 2] = u - v; // Butterfly operation: difference
                w *= wlen; // Move to the next root of unity
            }
        }
    }

    // If performing inverse FFT, scale the results by 1/n
    if (invert) {
        for (cld& x : a) {
            x /= n;
        }
    }
}


/**
 * @brief Multiplies two polynomials represented by vectors of coefficients using FFT.
 * 
 * @param a Coefficients of the first polynomial. a[i] is coefficient of z^i.
 * @param b Coefficients of the second polynomial. b[i] is coefficient of z^i.
 * @return Vector of coefficients of the product polynomial.
 * 
 * Handles potential zero polynomials. Uses FFT for efficiency. Rounds results to nearest long long.
 */
vector<ll> multiply(const vector<ll>& a, const vector<ll>& b) {
    // Check for zero polynomial inputs to optimize or handle edge cases
    bool a_is_zero = all_of(a.begin(), a.end(), [](ll v){ return v == 0; });
    bool b_is_zero = all_of(b.begin(), b.end(), [](ll v){ return v == 0; });
    
    // If either polynomial is zero, the product is the zero polynomial. Return [0].
    if (a_is_zero || b_is_zero) {
        return vector<ll>{0}; 
    }

    // Convert coefficient vectors to complex vectors for FFT
    vector<cld> fa(a.size()), fb(b.size());
    for(size_t i=0; i<a.size(); ++i) fa[i] = a[i];
    for(size_t i=0; i<b.size(); ++i) fb[i] = b[i];
    
    // Determine required FFT size: smallest power of 2 >= degree of product + 1
    int n = 1;
    // Degree of product = (a.size()-1) + (b.size()-1). Size needed = degree + 1 = a.size()+b.size()-1
    while (n < a.size() + b.size() - 1) n <<= 1; 
    
    // Resize vectors to FFT size, padding with zeros
    fa.resize(n);
    fb.resize(n);

    // Perform forward FFT on both polynomials
    fft_iterative(fa, false);
    fft_iterative(fb, false);
    
    // Pointwise multiplication in frequency domain
    for (int i = 0; i < n; i++)
        fa[i] *= fb[i];
    
    // Perform inverse FFT to get coefficients of the product polynomial
    fft_iterative(fa, true);

    // Maximum possible degree of the product polynomial
    // If a has degree deg_a = a.size()-1, b has deg_b = b.size()-1, product has degree deg_a + deg_b.
    // Minimum degree is 0 if both input polynomials have degree 0.
    int res_max_deg = (a.size() > 0 ? a.size() - 1 : 0) + (b.size() > 0 ? b.size() - 1 : 0);
    int res_size = res_max_deg + 1; // Size of result vector based on max possible degree
    
    vector<ll> result(res_size); 
    // Extract real parts and round to nearest long long for coefficients
    for (int i = 0; i < res_size; i++) {
         // Check index validity against FFT result size `n`
         if (i < fa.size()) {
             // Round the real part to get the integer coefficient
             result[i] = ROUND_FUNC(fa[i].real()); 
         } else {
             // Coefficients beyond n are effectively zero
             result[i] = 0; 
         }
    }

    // Optional: Trim trailing zeros from the result vector
    // This makes the vector represent the exact degree of the polynomial
     int true_size = result.size();
     while (true_size > 1 && result[true_size - 1] == 0) {
         true_size--;
     }
     result.resize(true_size);

    // Ensure the result vector is never empty unless it represents the zero polynomial ([0])
    if (result.empty()) return vector<ll>{0};

    return result;
}


int main() {
    // Optimize input/output operations
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int N; // Number of slimes
    int M; // Number of bonus conditions
    ll X;  // Score per eaten slime
    cin >> N >> M >> X;

    // Handle the edge case N=0 immediately
    if (N == 0) { 
        cout << 0 << endl;
        return 0;
    }

    // Read slime colors (1-based indexing in problem, stored 0-based in vector C)
    vector<int> C(N);
    for (int i = 0; i < N; ++i) {
        cin >> C[i];
    }

    // Precompute P_c polynomials for each color c. P_coeffs[c][a] = total bonus Y_j for color c at position a.
    vector<vector<ll>> P_coeffs(6, vector<ll>(N + 1, 0)); // Size N+1 for degrees 0 to N. Index 0 unused for positions.
    for (int j = 0; j < M; ++j) {
        int A; // Required position (1-based)
        int B; // Required color (1 to 5)
        ll Y;  // Bonus score
        cin >> A >> B >> Y;
        // Validate input bounds before using them as indices
        if (A >= 1 && A <= N && B >= 1 && B <= 5) { 
             P_coeffs[B][A] += Y; // Add bonus Y to coefficient for z^A for color B
        }
    }

    // Initialize total scores for eating k slimes (k=0 to N). Initial score is k*X.
    vector<ll> total_score(N + 1);
    for (int k = 0; k <= N; ++k) {
        total_score[k] = (ll)k * X;
    }

    // For each color c from 1 to 5
    for (int c = 1; c <= 5; ++c) {
        // Copy P_coeffs for current color c to avoid modification if needed elsewhere
        vector<ll> current_P_coeffs = P_coeffs[c]; 
        
        // Construct T_c polynomial based on slime positions with color c.
        // T_coeffs[p] = 1 if there is a slime of color c at original position i such that p = N-i.
        vector<ll> current_T_coeffs(N + 1, 0); // Size N+1 for degrees 0 to N.
        bool color_exists = false; // Flag to check if any slime has color c
        for (int i = 0; i < N; ++i) { // Iterate through 0-based indices
            if (C[i] == c) {
                // Slime at original position i+1 has color c
                // Set coefficient for z^{N-(i+1)} to 1. Index is N-(i+1).
                current_T_coeffs[N - (i + 1)] = 1; 
                color_exists = true;
            }
        }
        
        // Check if P or T polynomials are effectively zero before FFT multiplication
        bool P_is_zero = all_of(current_P_coeffs.begin(), current_P_coeffs.end(), [](ll v){ return v == 0; });
        
        // If no slime of color c exists, T_c is zero. If no bonuses target color c, P_c is zero.
        // If either is zero, their product is zero, so skip FFT.
        if (!color_exists || P_is_zero) {
            continue; 
        }

        // Compute the product polynomial Q_c = P_c * T_c using FFT
        vector<ll> Q_c = multiply(current_P_coeffs, current_T_coeffs);

        // Add the contribution of Q_c coefficients to the total score
        // The coefficient Q_c[N-k] corresponds to the bonus points obtained when eating k slimes for color c.
        for (int k = 0; k <= N; ++k) {
            int exponent = N - k; // Exponent relevant for k eaten slimes
            // Check if the exponent is a valid index in the result polynomial Q_c
            if (exponent >= 0 && exponent < Q_c.size()) {
                total_score[k] += Q_c[exponent]; // Add bonus points to score for eating k slimes
            }
        }
    }

    // Find the maximum score among all possible numbers of eaten slimes (k=0 to N)
    ll max_score = 0;
     if (!total_score.empty()) { // Should not be empty due to N=0 check earlier
       max_score = total_score[0]; // Initialize max score with the score for k=0
       // Iterate from k=1 to N to find the overall maximum score
       for (int k = 1; k <= N; ++k) {
           // Check index validity (though total_score has size N+1)
           if (k < total_score.size()) { 
              max_score = max(max_score, total_score[k]);
           }
       }
       // Ensure the final maximum score is non-negative (as per problem constraints X>=0, Y>=1)
       max_score = max(0LL, max_score); 
    } 

    // Output the maximum possible score
    cout << max_score << endl;

    return 0;
}
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