"Unique Object Codes for Inter-Group Connections"

Language: C++

Author: Jerry Obrian

Comments: 0

Useful:

Code:

We are given N, and we know there are 1..N objects in group 1 and N+1...2*N objects in group 2. Every object in group 1 is connected with each other, and every object in group 2 is connected with each other (inside group). There is a list consisting of length p (1 <= p <= N^2) of numbers [a, b], where a is object from group 1 and b is object from group 2. The pair [a, b] means that there must be connection between object a and b. If there is no [a, b] pair in the list, then there cannot be connection between them. For every object, we need to set unique code representing this object, which is string constisting of characters {A, B, C} with length up to M and N + 1 <= M <= 3N. Given two codes x and y of equal length, if there is index i such that x_i == y_i, then there is connection between object represented by code x and y. I tried solving it starting with setting characters in codes that will create required connections between objects of two distinct groups. #include <iostream> #include <vector> #include <set> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); int n, p, M; cin >> n >> p >> M; set<int> groups; set<pair<int, int>> connections; for (int i = 0; i < p; ++i) { int a, b; cin >> a >> b; groups.insert(b); connections.insert({a, b}); } vector<string> codes(2 * n + 1, string(M, ' ')); int pos = 1; set<int> a_pos; for (const int b : groups) { for (int a = 1; a <= n; ++a) { if (connections.find({a, b}) == connections.end()) continue; codes[a][pos] = 'A'; codes[b][pos] = 'A'; } ++pos; a_pos.insert(pos); } }

Details:

To use as minimum characters as possible to create required connections, I grouped connections basing on their endpoints (objects from group 2). This way, assume N = 3, if we have connections e.g. 1 -> 4, 2 -> 4, we can use one character at some position to connect both 1 and 2 to 4. Also, I assumed my codes will always be of maximum possible length. However, I have problem to come up with logic that will ensure uniqueness of every code. We can see that if we imagine the connections between objects as a graph, then objects become nodes, there are two main components (group 1 and group 2), and then we have connections between these components, so it's a bit similar to a bipartite graph, but this observation didn't help me. How can we solve that? That's problem Satelity from XXXI Polish Olympiad in Informatics, stage I

Solution:

To ensure the uniqueness of every code representing an object as explained, you need to think of the connections forming a bipartite graph where you have group 1 nodes connected to group 2 nodes. In this scenario, you can think of the objects within the same group as being equivalent, and you need to guarantee that by assigning unique codes to these objects.

Here's a suggested approach to guarantee uniqueness:

1. **Create a Unique Character Mapping**: For each object `x` in group 1, you can assign a unique character `C_x` to it. Similarly, for each object `y` in group 2, you can assign a unique character `C_y` to it.

2. **Character Building Logic**:
- For each connection [a, b]:
- Set `codes[a][b] = C_b` and `codes[b][b] = C_b`.

3. **Filling up the Rest of the Code**:
- For each position `pos` greater than the number of connections, assign a unique character to ensure uniqueness within each group.

Here is an updated version of your code reflecting these changes:

cpp
#include 
#include 
#include 

using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int n, p, M;
    cin >> n >> p >> M;

    // Create a unique character mapping for each object
    vector group1Char(n + 1);
    vector group2Char(n + 1);
    for (int i = 1; i <= n; ++i) {
        group1Char[i] = 'A' + (i - 1);
        group2Char[i] = 'A' + (i - 1);
    }

    set> connections;
    for (int i = 0; i < p; ++i)
    {
        int a, b;
        cin >> a >> b;

        connections.insert({a, b});
    }

    // Initialize codes with a unique character for each object
    vector codes(2 * n + 1, string(M, ' '));
    for (int a = 1; a <= n; ++a)
    {
        codes[a][b] = group1Char[a];  // Assign unique character for the object in group 1
        codes[a + n][b] = group2Char[a];  // Assign unique character for the object in group 2
    }

    // Set characters in codes to create required connections
    int pos = n + 1;
    for (auto [a, b] : connections)
    {
        codes[a][pos] = group2Char[b];  // Connect object a with b in group 2
        codes[b + n][pos] = group2Char[b];  // Connect object b with a in group 1
    }

    // Assign unique characters for the rest of the positions
    for (int i = pos + 1; i <= M; ++i)
    {
        codes[i][i] = 'A' + n + 1 + (i - pos);  // Assign unique characters for unmatched positions
    }

    // Output the codes
    for (const string& code : codes)
    {
        cout << code << endl;
    }

    return 0;
}

This logic should ensure the uniqueness of every code while minimizing the number of characters needed to create the required connections.

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