Problem Analysis

The road construction problem involves determining the minimum time required to complete paving operations across multiple segments. While algorithm tags suggest greedy approaches and binary indexed trees, a dynamic programming solution provides an elegant and efficeint implementation.

Understanding the Solution Pattern

Consider the input sequence [5, 1, 2, 3]. Each segment's completion time depends on its relationship with preceding segments:

• If current depth is less than or equal to previous maximum, it inherits the earlier completion time
• If current depth exceeds the previous maximum, additional time equals thier difference

Initial Rceursive Approach

A divide-and-conquer strategy was initially attempted, partitioning the array into intervals and processing minimum values recursively:

#include <bits/stdc++.h>
using namespace std;

int total_time = 0;
int segment_count = 0;

void find_minimum(int data[], int& min_val, int& pos, int start, int end) {
    int lowest = INT_MAX;
    for (int idx = start; idx <= end; idx++) {
        if (data[idx] < lowest) {
            lowest = data[idx];
            pos = idx;
        }
        if (data[idx] == 0) {
            pos = idx;
            return;
        }
    }
    min_val = lowest;
}

bool all_zero(int start, int end, int data[]) {
    for (int idx = start; idx <= end; idx++) {
        if (data[idx] != 0) {
            return false;
        }
    }
    return true;
}

void solve_recursive(int start, int end, int data[]) {
    if (all_zero(start, end, data)) {
        return;
    }
    
    int min_value = 0;
    int min_position = 0;
    find_minimum(data, min_value, min_position, start, end);
    
    total_time += min_value;
    for (int idx = start; idx <= end; idx++) {
        data[idx] -= min_value;
    }
    
    solve_recursive(start, min_position - 1, data);
    solve_recursive(min_position + 1, end, data);
}

int main() {
    scanf("%d", &segment_count);
    int* array = (int*)malloc(segment_count * sizeof(int));
    
    for (int i = 0; i < segment_count; i++) {
        scanf("%d", &array[i]);
    }
    
    solve_recursive(0, segment_count - 1, array);
    printf("%d", total_time);
    
    free(array);
    return 0;
}

Optimized Dynamic Programming Solution

The dynamic programming approach simplifies the calculation by tracking cumulative effects from left to right:

#include <bits/stdc++.h>
using namespace std;

int main() {
    int count;
    scanf("%d", &count);
    
    int* depths = (int*)malloc(count * sizeof(int));
    int* cumulative_times = (int*)malloc(count * sizeof(int));
    
    for (int i = 0; i < count; i++) {
        scanf("%d", &depths[i]);
        
        if (i == 0) {
            cumulative_times[i] = depths[i];
        } else {
            if (depths[i] <= depths[i - 1]) {
                cumulative_times[i] = cumulative_times[i - 1];
            } else {
                cumulative_times[i] = cumulative_times[i - 1] + depths[i] - depths[i - 1];
            }
        }
    }
    
    printf("%d", cumulative_times[count - 1]);
    
    free(depths);
    free(cumulative_times);
    return 0;
}

Algorithm Logic

Each position's value reflects the influence of all preceding positions. When a segment requires more time than its predecessor, the additional time accumulates. When it requires less time, it can be completed within the previous segment's timeframe due to overlapping work capabilities.