CPMagical CP Library

Problem solution · C++

ABC402 C — Dislike Foods

ABC402 C — Dislike Foods: a C++ solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to michimani AtCoder Solutions.

Technique
Hash-based lookup
Source
michimani AtCoder Solutions
Length
41 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

Hash-based lookup

For ABC402 C — Dislike Foods, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 41 lines of C++ from the credited upstream file abc402_c.cpp.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 4 loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeABC402 C — Dislike Foods · C++C++
Use this to learn the idea, then write your own version.
#include <iostream>#include <unordered_map>#include <vector> using namespace std;using ui = unsigned int; int main() {    ui n, m;    cin >> n >> m;     vector<ui> mk(m);     unordered_map<ui, vector<ui>> nm;     ui k, a;    for (ui i = 0; i < m; i++) {        cin >> k;        mk[i] = k;        for (ui j = 0; j < k; j++) {            cin >> a;            nm[a].push_back(i);        }    }     ui c = 0;     ui food;    for (ui day = 0; day < n; day++) {        cin >> food;         auto it = nm.find(food);        if (it != nm.end())            for (ui dish_id : it->second)                if (--mk[dish_id] == 0) c++;         cout << c << 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 ↗