C++ Coding Test Course, Making Travel Plans

Problem Solving Process

1. Problem Analysis

This problem can be solved using graph traversal and dynamic programming. Each city can be considered a vertex and the costs between cities as edges. To visit as many cities as possible within the given budget, we will utilize backtracking (DFS) to explore all possible routes. A cost of -1 indicates a city that cannot be traveled to, so these cases need to be excluded.

2. Algorithm Design

The basic algorithm to solve the problem is as follows:

1. Declare variables to track the number of visited cities and the remaining budget from the current city.
2. Implement depth-first search (DFS) to recursively explore all cities reachable from the current city.
3. Update the remaining budget and the count of visited cities when returning to the parent node.
4. After exploring all paths of the recursive calls, compare and update the maximum visited cities count.

3. C++ Code Implementation

Below is an example implementation of the above algorithm in C++:

#include <iostream> 
#include <vector> 
#include <algorithm> 

using namespace std; 

int N, B; 
vector<vector<int>> cost; 
int maxCities = 0; 

void dfs(int currentCity, int currentBudget, int visitedCount) { 
    maxCities = max(maxCities, visitedCount); 

    for (int nextCity = 0; nextCity < N; nextCity++) { 
        if (cost[currentCity][nextCity] != -1 && currentBudget - cost[currentCity][nextCity] >= 0) { 
            // Move to the next city 
            dfs(nextCity, currentBudget - cost[currentCity][nextCity], visitedCount + 1); 
        } 
    } 
} 

int main() { 
    cout << "Enter the number of locations N: "; 
    cin >> N; 
    cost.resize(N, vector<int>(N)); 

    cout << "Enter the cost matrix: " << endl; 
    for (int i = 0; i < N; i++) { 
        for (int j = 0; j < N; j++) { 
            cin >> cost[i][j]; 
        } 
    } 

    cout << "Enter the total budget B: "; 
    cin >> B; 

    for (int i = 0; i < N; i++) { 
        dfs(i, B, 1); // Start DFS from each city 
    } 

    cout << "Maximum number of cities that can be visited: " << maxCities << endl; 
    return 0; 
} 

4. Code Explanation

The above code takes the travel costs between cities as input and performs depth-first search starting from each city.

• maxCities: A variable to store the maximum number of cities visited.
• dfs function: Takes the current city, remaining budget, and number of visited cities as parameters to explore all possible routes.
• The main function receives input from the user and calls DFS from all cities to find the maximum number of cities that can be visited.

5. Execution Result

When the code is compiled and executed, the following output can be obtained:

Enter the number of locations N: 4
Enter the cost matrix: 
0 10 15 -1
10 0 35 20
15 35 0 30
-1 20 30 0
Enter the total budget B: 50
Maximum number of cities that can be visited: 3

6. Conclusion

In this lecture, we discussed an algorithm problem related to planning a trip using C++. We learned the process of calculating the maximum number of cities that can be visited within a budget using depth-first search (DFS) and backtracking. This helped us understand traversal algorithms and how they can be applied to solve real-world problems.

Backtracking techniques are not limited to this problem and can be applied to various combination and permutation problems, so it is important to understand the fundamental concepts of algorithms accurately. It would be beneficial to practice more algorithm problems in the future to gain more experience.