C++ Coding Test Course, Bridge Building

Problem Definition

The bridge placement problem is to find the number of bridges that can be placed between poles when N bridges and M poles are given. The bridges placed between each pole must be connectable, and only one bridge can be placed on each pole. This problem can be solved using combinatorial concepts and dynamic programming.

Problem Example

For example, let’s consider the case when N = 2 and M = 4. When there are poles A, B, C, and D as shown below, the ways to place the bridges are as follows:

A – B
A – C
A – D
B – C
B – D
C – D

A maximum of 6 combinations is possible as shown above.

Understanding and Analyzing the Problem

To solve the problem, the concept of combinations must be applied. This problem is about choosing N bridges from M poles, and the number of combinations can be expressed with the following formula.

            C(M, N) = M! / (N! * (M - N)!)

Here, C represents combinations, M is the total number of poles, and N is the number of bridges to be placed. When implementing this in C++, a function to calculate combinations must be created.

Algorithm Design

Depending on the given values of N and M, the bridge placement problem can be solved by determining the arrangement of bridges using combinations. The recurrence relation for the bridge placement problem is as follows:

            dp[n][m] = dp[n-1][m-1] + dp[n][m-1]

Here, dp[n][m] indicates the number of ways to place n bridges on m poles. It can be calculated by adding the number of ways to place a bridge on one pole and the number of ways to skip placing a bridge.

Algorithm Implementation

Below is an example code implementing this algorithm in C++:

#include <iostream>
#include <vector>

using namespace std;

long long combinations(int n, int r) {
    if (r > n) return 0;
    if (r == 0 || r == n) return 1;
    vector<long long> dp(n + 1);
    dp[0] = 1;

    for (int i = 1; i <= n; ++i) {
        for (int j = i; j >= 1; --j) {
            dp[j] += dp[j - 1];
        }
    }
    return dp[r];
}

int main() {
    int n, m;
    cout << "Enter the number of bridges and the number of poles: ";
    cin >> n >> m;

    long long result = combinations(m, n);
    cout << "Maximum number of possible combinations for bridge placement: " << result << endl;
    return 0;
}

Code Explanation

In the above code, we take input for the number of bridges and the number of poles, and calculate the combinations to output the result. The combinations function uses dynamic programming to compute the combinations. This allows for efficient calculation.

Complexity Analysis

The time complexity of this algorithm is O(n * r), where r is the number of bridges selected from poles. The space complexity is also O(n). Thanks to this efficient structure, results can be derived quickly even for considerably large values.

Conclusion

The bridge placement problem is a good exercise that can be efficiently solved by understanding the basics of combinatorics and utilizing dynamic programming. This method is useful not only in coding tests but also in solving various combinatorial problems. Through this problem, you will be able to enhance both your basic programming skills and your problem-solving abilities.