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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:

In the provided code, the issue seems to be related to ensuring the uniqueness of the codes for each object. The current approach of assigning characters 'A' to the codes based on connections is a good start, but to ensure uniqueness, you need to handle collisions where a character is already assigned to another object.

Here is an approach you can follow:

1. Start by initializing a set to keep track of the characters already assigned to each position in the code.
2. For each connection [a, b], if we need to assign 'A' to both a and b at position pos, check if 'A' is already assigned to any other object at position pos.
3. If 'A' is already assigned to some other object at position pos, try assigning 'B' or 'C' at position pos for a or b. Ensure this new assignment is not conflicting with any other already assigned characters at that position.
4. Update the set of assigned characters for the current position accordingly.

You can implement this by modifying the existing code like this:

cpp
// Initialize a set to keep track of assigned characters at each position
vector> assigned_chars(M, set());

// Modify the assignment logic to handle uniqueness
for (const int b : groups)
{
    for (int a = 1; a <= n; ++a)
    {
        if (connections.find({a, b}) == connections.end())
            continue;

        char assigned_char = 'A';
        if (assigned_chars[pos].count('A') > 0)
        {
            assigned_char = (assigned_chars[pos].count('B') == 0) ? 'B' : 'C';
        }

        codes[a][pos] = assigned_char;
        codes[b][pos] = assigned_char;

        // Update the set of assigned characters for the current position
        assigned_chars[pos].insert(assigned_char);
    }

    ++pos;
}

This modification will help in ensuring the uniqueness of the codes for each object while minimizing the characters used to create the required connections.

"Unique Codes for Object Groups Connection"