CPMagical CP Library

Problem solution · C++

ABC415 C — Mixture

ABC415 C — Mixture: a C++ solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to michimani AtCoder Solutions.

Technique
Breadth-first search
Source
michimani AtCoder Solutions
Length
48 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For ABC415 C — Mixture, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 48 lines of C++ from the credited upstream file abc415_c.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 3 loop blocks detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from michimani AtCoder Solutions by michimani and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeABC415 C — Mixture · C++C++
Use this to learn the idea, then write your own version.
#include <iostream>#include <queue>#include <vector> using namespace std;using ui = unsigned int; bool can_mix_all(ui n, string s) {    vector<bool> visited(1 << n, false);    queue<int> q;     q.push(0);    visited[0] = true;     while (!q.empty()) {        int current = q.front();        q.pop();         if (current == (1 << n) - 1) return true;         for (ui i = 0; i < n; i++) {            if (!(current & (1 << i))) {                int next = current | (1 << i);                if (!visited[next] && s[next - 1] == '0') {                    visited[next] = true;                    q.push(next);                }            }        }    }     return false;} int main() {    ui t;    cin >> t;     for (; t--;) {        ui n;        string s;        cin >> n >> s;         cout << (can_mix_all(n, s) ? "Yes" : "No") << endl;    }     return 0;}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗