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Problem solution · C++

ABC325 C — Sensors

ABC325 C — Sensors: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to michimani AtCoder Solutions.

Technique
Direct simulation
Source
michimani AtCoder Solutions
Length
66 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

Direct simulation

For ABC325 C — Sensors, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 66 lines of C++ from the credited upstream file abc325_c.cpp.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 codeABC325 C — Sensors · C++C++
Use this to learn the idea, then write your own version.
#include <algorithm>#include <iostream>#include <vector> using namespace std; void sensor_traverse(vector<string>& grid, unsigned long i, unsigned long j,                     vector<string>& visited, int& visited_sensor_count) {    visited[i][j] = '#';    visited_sensor_count++;    for (unsigned long ti = max((unsigned long)(1), i - 1);         ti <= min(i + 1, (unsigned long)(grid.size() - 1)); ti++) {        for (unsigned long tj = max((unsigned long)(1), j - 1);             tj <= min(j + 1, (unsigned long)(grid[0].size())); tj++) {            if (ti == i && tj == j) {                continue;            }             if (visited[ti][tj] != '#' && grid[ti][tj] == '#') {                sensor_traverse(grid, ti, tj, visited, visited_sensor_count);            }        }    }} int main() {    unsigned long h, w;    cin >> h >> w;     string dummy(w + 2, '.');    vector<string> grid(h + 2, dummy);    int sensor_count = 0;    for (unsigned long j = 1; j <= h; j++) {        string s;        cin >> s;        sensor_count += count(s.begin(), s.end(), '#');        s = "." + s + ".";        grid[j] = s;    }     int ans = 0;    vector<string> visited(h + 2, dummy);    int visited_sensor_count = 0;    for (unsigned long i = 1; i <= h; i++) {        if (visited_sensor_count == sensor_count) {            cout << ans << endl;            return 0;        }         for (unsigned long j = 1; j <= w; j++) {            if (visited_sensor_count == sensor_count) {                cout << ans << endl;                return 0;            }             if (grid[i][j] != '#' || visited[i][j] == '#') {                continue;            }             ans++;            sensor_traverse(grid, i, j, visited, visited_sensor_count);        }    }     cout << ans << endl;}

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