题解思路

这是一道典型的二分答案+贪心策略验证问题，核心在于将最小化最大值问题转化为判定性问题，通过双向约束传播实现最优解。

核心算法分析

问题转换思路： 原问题要求最小化相邻差的最大值，这是经典的最小化最大值优化问题。我们可以转换为判定问题：给定阈值X，能否通过k次操作使所有相邻差都不超过X？

关键洞察： 对于给定的X，要使|a[i+1] - a[i]| ≤ X，等价于： a[i]−X≤a[i+1]≤a[i]+Xa[i] - X \leq a[i+1] \leq a[i] + X

双向约束传播策略：

前向传播：确保右侧元素满足左侧约束

for i in range(1, n):
    if b[i] < b[i-1] - X:
        b[i] = b[i-1] - X  // 右侧至少要达到这个下界

后向传播：确保左侧元素满足右侧约束

for i in range(n-2, -1, -1):
    if b[i] < b[i+1] - X:
        b[i] = b[i+1] - X  // 左侧至少要达到这个下界

算法正确性证明：

1. 单调性：X越大，约束越宽松，所需操作次数单调递减
2. 最优性：双向传播确保了在满足约束条件下的最小增长
3. 完整性：两次扫描保证了所有相邻约束都被满足

详细算法设计

二分搜索框架：

private static boolean canAchieve(int[] array, int maxDiff, long maxOps) {
    int n = array.length;
    long[] modified = new long[n];
    
    // 复制原数组（使用long防止溢出）
    for (int i = 0; i < n; i++) {
        modified[i] = array[i];
    }
    
    // 前向约束传播
    for (int i = 1; i < n; i++) {
        if (modified[i] < modified[i-1] - maxDiff) {
            modified[i] = modified[i-1] - maxDiff;
        }
    }
    
    // 后向约束传播
    for (int i = n-2; i >= 0; i--) {
        if (modified[i] < modified[i+1] - maxDiff) {
            modified[i] = modified[i+1] - maxDiff;
        }
    }
    
    // 计算总操作次数
    long totalOps = 0;
    for (int i = 0; i < n; i++) {
        totalOps += modified[i] - array[i];
        if (totalOps > maxOps) return false; // 提前终止
    }
    
    return true;
}

复杂度分析：

• 单次验证：O(n)
• 二分搜索：O(log(max_diff))，其中max_diff为初始最大相邻差
• 总体复杂度：O(n log(max_diff))
• 空间复杂度：O(n)

Java代码实现

import java.io.*;
import java.util.*;

public class Main {
    
    private static long calculateRequiredOperations(int[] originalArray, int targetMaxDiff) {
        int length = originalArray.length;
        long[] adjustedArray = new long[length];
        
        // 初始化调整后的数组
        for (int i = 0; i < length; i++) {
            adjustedArray[i] = originalArray[i];
        }
        
        // 前向传播：保证右侧元素满足左侧约束
        for (int i = 1; i < length; i++) {
            long minimumRequired = adjustedArray[i-1] - targetMaxDiff;
            if (adjustedArray[i] < minimumRequired) {
                adjustedArray[i] = minimumRequired;
            }
        }
        
        // 后向传播：保证左侧元素满足右侧约束
        for (int i = length - 2; i >= 0; i--) {
            long minimumRequired = adjustedArray[i+1] - targetMaxDiff;
            if (adjustedArray[i] < minimumRequired) {
                adjustedArray[i] = minimumRequired;
            }
        }
        
        // 计算所需的总操作次数
        long totalOperations = 0;
        for (int i = 0; i < length; i++) {
            totalOperations += adjustedArray[i] - originalArray[i];
        }
        
        return totalOperations;
    }
    
    private static boolean isAchievable(int[] array, int maxDifferenceLimit, long availableOperations) {
        return calculateRequiredOperations(array, maxDifferenceLimit) <= availableOperations;
    }
    
    private static int findMinimumMaxDifference(int[] array, long maxOperations) {
        int n = array.length;
        
        // 确定二分搜索的右边界：初始最大相邻差
        int rightBound = 0;
        for (int i = 0; i < n - 1; i++) {
            int currentDiff = Math.abs(array[i+1] - array[i]);
            rightBound = Math.max(rightBound, currentDiff);
        }
        
        int leftBound = 0;
        
        // 二分搜索最小可能的最大相邻差
        while (leftBound < rightBound) {
            int midValue = leftBound + (rightBound - leftBound) / 2;
            
            if (isAchievable(array, midValue, maxOperations)) {
                rightBound = midValue;
            } else {
                leftBound = midValue + 1;
            }
        }
        
        return leftBound;
    }
    
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int testCases = Integer.parseInt(reader.readLine());
        
        StringBuilder resultBuilder = new StringBuilder();
        
        for (int t = 0; t < testCases; t++) {
            String[] parameters = reader.readLine().split(" ");
            int n = Integer.parseInt(parameters[0]);
            long k = Long.parseLong(parameters[1]);
            
            String[] arrayElements = reader.readLine().split(" ");
            int[] inputArray = new int[n];
            
            for (int i = 0; i < n; i++) {
                inputArray[i] = Integer.parseInt(arrayElements[i]);
            }
            
            int result = findMinimumMaxDifference(inputArray, k);
            resultBuilder.append(result).append('\n');
        }
        
        System.out.print(resultBuilder.toString());
        reader.close();
    }
}

C++代码实现

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;

class AdjacentDifferenceOptimizer {
private:
    static long long computeMinOperations(const vector<int>& original, int maxAllowedDiff) {
        int size = original.size();
        vector<long long> modified(size);
        
        // 复制原始数组到long long类型以防溢出
        for (int i = 0; i < size; i++) {
            modified[i] = original[i];
        }
        
        // 正向扫描：确保每个元素满足左邻居的约束
        for (int i = 1; i < size; i++) {
            long long lowerBound = modified[i-1] - maxAllowedDiff;
            if (modified[i] < lowerBound) {
                modified[i] = lowerBound;
            }
        }
        
        // 反向扫描：确保每个元素满足右邻居的约束
        for (int i = size - 2; i >= 0; i--) {
            long long lowerBound = modified[i+1] - maxAllowedDiff;
            if (modified[i] < lowerBound) {
                modified[i] = lowerBound;
            }
        }
        
        // 统计总的增量操作
        long long totalIncrements = 0;
        for (int i = 0; i < size; i++) {
            totalIncrements += modified[i] - original[i];
        }
        
        return totalIncrements;
    }
    
public:
    static int solveMinMaxDifference(const vector<int>& array, long long maxOperations) {
        int arraySize = array.size();
        
        // 计算初始状态下的最大相邻差作为搜索上界
        int searchUpperBound = 0;
        for (int i = 0; i < arraySize - 1; i++) {
            int adjacentDiff = abs(array[i+1] - array[i]);
            searchUpperBound = max(searchUpperBound, adjacentDiff);
        }
        
        int searchLowerBound = 0;
        
        // 使用二分搜索寻找最优解
        while (searchLowerBound < searchUpperBound) {
            int candidateAnswer = searchLowerBound + (searchUpperBound - searchLowerBound) / 2;
            
            long long requiredOps = computeMinOperations(array, candidateAnswer);
            
            if (requiredOps <= maxOperations) {
                searchUpperBound = candidateAnswer;
            } else {
                searchLowerBound = candidateAnswer + 1;
            }
        }
        
        return searchLowerBound;
    }
};

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int numberOfTests;
    cin >> numberOfTests;
    
    while (numberOfTests--) {
        int n;
        long long k;
        cin >> n >> k;
        
        vector<int> inputSequence(n);
        for (int i = 0; i < n; i++) {
            cin >> inputSequence[i];
        }
        
        int optimalResult = AdjacentDifferenceOptimizer::solveMinMaxDifference(inputSequence, k);
        cout << optimalResult << '\n';
    }
    
    return 0;
}

时间复杂度：O(n log(max_diff))，其中max_diff为初始最大相邻差 空间复杂度：O(n)，用于存储临时调整数组