Problem Overview

The problem involves merging a sequence of stones where each stone has a weight. The goal is to merge consecutive stones within a sequence into a single stone with a weight equal to the sum of the merged stones, at a cost equal to that sum. The merging must result in a final number of stones between a given range [L, R], and the objective is to minimize the total cost.

Dynamic Programming Approach

A straightforward dynamic programming solution can be formulated. Let dp[l][r][d] represent the minimum cost to merge the subsequence from endex l to r into exactly d stones. For d > 1, after computing the DP values, they can contribute to the case where d = 1.

This approach has a time complexity of O(n⁴), which might be acceptable with optimizations but could be borderline for larger constraints.

Optimized Implementation

To improve performance, we can use rolling arrays for the same starting index l and swap dimensions to ensure memory access patterns are cache-friendly.

#include <cstdio>
#include <climits>
using namespace std;

const int MAX_N = 403;
const int INF = INT_MAX;

int mergeCost[MAX_N][MAX_N], tempDP[MAX_N][MAX_N];

inline void updateMin(int &current, int candidate) {
    if (current > candidate) current = candidate;
}

int readInt() {
    int value = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9') {
        value = value * 10 + (ch - '0');
        ch = getchar();
    }
    return value;
}

int prefixSum[MAX_N];

int main() {
    int testCases = readInt();
    while (testCases--) {
        int n = readInt();
        int minParts = readInt();
        int maxParts = readInt();
        prefixSum[0] = 0;
        for (int i = 1; i <= n; ++i) {
            prefixSum[i] = readInt() + prefixSum[i - 1];
        }

        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= n; ++j) {
                mergeCost[i][j] = tempDP[i][j] = INF;
            }
        }

        for (int left = n; left >= 1; --left) {
            tempDP[left][left] = 0;
            mergeCost[1][left] = 0;
            for (int right = left + 1; right <= n; ++right) {
                tempDP[right][left] = INF;
                for (int parts = 2; parts <= right - left + 1 && parts <= maxParts; ++parts) {
                    int best = INF;
                    for (int split = left + parts - 2; split < right; ++split) {
                        if (mergeCost[parts - 1][split] != INF && tempDP[right][split + 1] != INF) {
                            updateMin(best, mergeCost[parts - 1][split] + tempDP[right][split + 1]);
                        }
                    }
                    if (minParts <= parts && parts <= maxParts && best != INF) {
                        updateMin(tempDP[right][left], best + prefixSum[right] - prefixSum[left - 1]);
                    }
                    mergeCost[parts][right] = best;
                }
                mergeCost[1][right] = tempDP[right][left];
            }
        }

        if (tempDP[n][1] == INF) {
            printf("0\n");
        } else {
            printf("%d\n", tempDP[n][1]);
        }
    }
    return 0;
}

Advanced Technique: Matrix Exponentiation

Consider a more generalized problem: given a cost matrix w[l][r] for each interval, find for every interval [l, r] the minimum total cost to partition it into x subintervals where x is in [L, R].

When L = R, this reduces to computing the L-th power of the cost matrix under the (min, +) semiring, which can be done efficciently using exponentiation by squaring.

For the range [L, R], we need the sum Σ_{i=L}^{R} Wⁱ, where addition is defined as element-wise minimum. This can be expressed as Wᴸ * Σ_{i=0}^{R-L} Wⁱ. The summation part can be computed using divide-and-conquer techniques.

However, in our specific problem, the cost w[l][r] depends on the DP state f[l][r][d], which in turn is derived from the matrix multiplication of w. This interdependence resembles "semi-online convolution," where we process intervals in increasing order of size.

Conceptual Framework

We can think of this as having an outer layer of matrix multiplication and an inner layer that performs a form of exponentiation. Instead of tracking the exact number of parts d, we use a structure akin to a "fast exponentiation automaton" (though not a true automaton).

Define:

• pow2Cost[l][r][t]: minimum cost to partition [l, r] into exactly 2ᵗ subintervals.
• rangeCost[l][r][t]: minimum cost to partition [l, r] into any number of subintervals from 1 to 2ᵗ.

Transitions can be derived from smaller intervals and powers.

Implementation with ST-Table Decomposition

A practical approach decomposes the range [L, R] into two overlapping intervals using binary representation, similar to an ST tible. This method, while not fully optimized, reduces complexity compared to the naive O(n⁴) solution.

#include <cstdio>
#include <algorithm>
#include <climits>
using namespace std;

typedef unsigned int uint;

uint readUInt() {
    uint value = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') ch = getchar();
    while (ch >= '0' && ch <= '9') {
        value = value * 10 + (ch - '0');
        ch = getchar();
    }
    return value;
}

const int MAX_N = 403;
const uint INF = 0x5fffffff;

inline void updateMin(uint &current, uint candidate) {
    if (current > candidate) current = candidate;
}

uint pow2Cost[MAX_N][MAX_N][9];      // Cost for exactly 2^t parts
uint rangeCost[MAX_N][MAX_N][9];      // Cost for 1 to 2^t parts
uint auxA[MAX_N][MAX_N][9], auxB[MAX_N][MAX_N][9]; // Auxiliary DP arrays
uint prefixSum[MAX_N];

int n, minParts, maxParts;
int bitsA[9], countA;
int bitsB[9], countB;

const char *NO_SOLUTION = "0";

int main() {
    int testCases = readUInt();
    while (testCases--) {
        n = readUInt();
        minParts = readUInt();
        maxParts = readUInt();
        for (int i = 1; i <= n; ++i) {
            prefixSum[i] = readUInt() + prefixSum[i - 1];
        }

        if (n == 1) {
            printf("0\n");
            continue;
        }
        if (minParts > maxParts) {
            puts(NO_SOLUTION);
            continue;
        }
        if (minParts == 1 && maxParts == 1) {
            puts(NO_SOLUTION);
            continue;
        }
        if (minParts == 1) ++minParts;

        if (minParts == maxParts) {
            int target = minParts;
            countA = 0;
            for (int bit = 0; bit < 9; ++bit) {
                if (target >> bit & 1) bitsA[countA++] = bit;
            }
            for (int length = 1; length <= n; ++length) {
                for (int left = 1, right = length; right <= n; ++left, ++right) {
                    for (int idx = 1; idx < countA; ++idx) {
                        auxA[left][right][idx] = INF;
                        for (int split = left; split < right; ++split) {
                            updateMin(auxA[left][right][idx], auxA[left][split][idx - 1] + pow2Cost[split + 1][right][bitsA[idx]]);
                        }
                    }
                    for (int bit = 1; bit < 9; ++bit) {
                        pow2Cost[left][right][bit] = INF;
                        for (int split = left; split < right; ++split) {
                            updateMin(pow2Cost[left][right][bit], pow2Cost[left][split][bit - 1] + pow2Cost[split + 1][right][bit - 1]);
                        }
                    }
                    pow2Cost[left][right][0] = INF;
                    if (left == right) pow2Cost[left][right][0] = 0;
                    auxA[left][right][0] = pow2Cost[left][right][bitsA[0]];
                    updateMin(pow2Cost[left][right][0], auxA[left][right][countA - 1] + prefixSum[right] - prefixSum[left - 1]);
                    updateMin(auxA[left][right][0], pow2Cost[left][right][bitsA[0]]);
                }
            }
        } else {
            int exponent = __lg(maxParts - minParts + 1);
            int lower = minParts - 1;
            int upper = maxParts - (1 << exponent);
            countA = countB = 0;
            for (int bit = 0; bit < 9; ++bit) {
                if (lower >> bit & 1) bitsA[countA++] = bit;
                if (upper >> bit & 1) bitsB[countB++] = bit;
            }
            for (int length = 1; length <= n; ++length) {
                for (int left = 1, right = length; right <= n; ++left, ++right) {
                    for (int idx = 1; idx < countA; ++idx) {
                        auxA[left][right][idx] = INF;
                        for (int split = left; split < right; ++split) {
                            updateMin(auxA[left][right][idx], auxA[left][split][idx - 1] + pow2Cost[split + 1][right][bitsA[idx]]);
                        }
                    }
                    for (int idx = 1; idx < countB; ++idx) {
                        auxB[left][right][idx] = INF;
                        for (int split = left; split < right; ++split) {
                            updateMin(auxB[left][right][idx], auxB[left][split][idx - 1] + pow2Cost[split + 1][right][bitsB[idx]]);
                        }
                    }
                    for (int bit = 1; bit < 9; ++bit) {
                        pow2Cost[left][right][bit] = INF;
                        for (int split = left; split < right; ++split) {
                            updateMin(pow2Cost[left][right][bit], pow2Cost[left][split][bit - 1] + pow2Cost[split + 1][right][bit - 1]);
                        }
                    }
                    pow2Cost[left][right][0] = INF;
                    for (int split = left; split < right; ++split) {
                        updateMin(pow2Cost[left][right][0], auxA[left][split][countA - 1] + rangeCost[split + 1][right][exponent] + prefixSum[right] - prefixSum[left - 1]);
                        updateMin(pow2Cost[left][right][0], auxB[left][split][countB - 1] + rangeCost[split + 1][right][exponent] + prefixSum[right] - prefixSum[left - 1]);
                    }
                    if (left == right) pow2Cost[left][right][0] = 0;
                    auxA[left][right][0] = pow2Cost[left][right][bitsA[0]];
                    auxB[left][right][0] = pow2Cost[left][right][bitsB[0]];
                    rangeCost[left][right][0] = pow2Cost[left][right][0];
                    for (int bit = 1; bit <= exponent; ++bit) {
                        rangeCost[left][right][bit] = pow2Cost[left][right][0];
                        for (int split = left; split < right; ++split) {
                            updateMin(rangeCost[left][right][bit], rangeCost[left][split][bit - 1] + rangeCost[split + 1][right][bit - 1]);
                        }
                    }
                }
            }
        }

        if (pow2Cost[1][n][0] == INF) {
            puts(NO_SOLUTION);
        } else {
            printf("%u\n", pow2Cost[1][n][0]);
        }
    }
    return 0;
}

Conclusion

This solution combines dynamic programming with matrix exponentiation concepts to efficiently solve the sequence merging problem. The first method provides a direct O(n⁴) DP with optimizations, while the second leverages binary decomposition and semi-online convolution to achieve better theoretical complexity, though with increased implementation complexity. The choice between them depends on specific constraints and performance requirements.