[Silver] 2D Conveyor Belt Debugging

Debugging Help
#1

Problem Info

USACO 2024 December Contest, Silver
Problem 3. 2D Conveyor Belt

Link: https://usaco.org/index.php?page=viewproblem2&cpid=1448

What I’ve Tried

I have checked the editorial and it aligns with my ideas:

  1. Find out the answer considered every query has been applied
  2. Process queries in reversed order
  3. For the current query, mark all other cells that point to the current cell (I used bfs)

I pass test cases 1, 2, 3, 6, and 10.

I have also tried using ChatGPT and it has been unable to debug my solution. I would appreciate any help!

My Work

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

vector<int> dx = {0, 0, 1, -1};
vector<int> dy = {1, -1, 0, 0};
vector<char> opposite_direction = {'L', 'R', 'U', 'D'};

struct Query {
    int r, c;
    char direction;
};

void printCoord(int x, int y) {
    cout << "(" << x << ", " << y << ")\n";
}

void bfs(int n, int& ans, queue<pair<int, int>>& q, vector<vector<bool>>& usable, vector<vector<char>>& grid) {
    while (!q.empty()) {
        auto cur = q.front(); q.pop();
        int x = cur.first; int y = cur.second;
        for (int k=0; k<4; k++) {
            int r = x + dx[k]; int c = y + dy[k];
            if (r > 0 && r <= n && c > 0 && c <= n) {
                if (!usable[r][c] && (grid[r][c] == opposite_direction[k] || grid[r][c] == '?')) {
                    usable[r][c] = true;
                    q.push({r, c});
                    ans++;
                }
            }
        }
    }
}

int main() {
    int n, num_queries; cin >> n >> num_queries;
    vector<vector<char>> grid(n+1, vector<char>(n+1, '?'));
    vector<vector<bool>> usable(n+1, vector<bool>(n+1, false));
    vector<Query> queries;
    for (int i=0; i<num_queries; i++) {
        int r, c; char d; cin >> r >> c >> d;
        Query cur; cur.r = r; cur.c = c; cur.direction = d;
        queries.push_back(cur);
        grid[r][c] = d;
    }
    int ans = 0;
    queue<pair<int, int>> q;
    for (int i=1; i<=n; i++) {
        if (grid[i][1] == '?' || grid[i][1] == 'L') {usable[i][1] = true;}
        if (grid[i][n] == '?' || grid[i][n] == 'R') {usable[i][n] = true;}
    }
    for (int j=1; j<=n; j++) {
        if (grid[1][j] == '?' || grid[1][j] == 'U') {usable[1][j] = true;}
        if (grid[n][j] == '?' || grid[n][j] == 'D') {usable[n][j] = true;}
    }
    for (int i=1; i<=n; i++) {
        for (int j=1; j<=n; j++) {
            if (usable[i][j]) {
                q.push({i, j});
            }
        }
    }
    bfs(n, ans, q, usable, grid);
    ans = 0;
    for (int i=1; i<=n; i++) {
        for (int j=1; j<=n; j++) {
            if (usable[i][j]) {
                ans++;
            }
        }
    }
    vector<int> answer;
    for (int i=num_queries-1; i>=0; i--) {
        answer.push_back(n*n - ans);
        int x = queries[i].r; int y = queries[i].c; char d = queries[i].direction;
        grid[x][y] = '?';
        if (!usable[x][y]) {
            for (int k=0; k<4; k++) {
                int r = x + dx[k]; int c = y + dy[k];
                if (r > 0 && r <= n && c > 0 && c <= n) {
                    if (usable[r][c] && (grid[r][c] == opposite_direction[k] || grid[r][c] == '?')) {
                        usable[x][y] = true;
                        ans++;
                        break;
                    }
                }
                else {
                    usable[x][y] = true;
                    ans++;
                    break;
                }
            }
        }
        if (usable[x][y]) {
            q.push({x, y});
            bfs(n, ans, q, usable, grid);
        }
    }
    for (int i=num_queries-1; i>=0; i--) {
        cout << answer[i] << '\n';
    }
}
#2

I believe your code is mostly correct, however, when you update the usability of a new cell after removing the conveyor belt, you make a check with ‘opposite_direction’. However, it does not matter whether the cells next to the new empty cell can reach the current cell, because as long as the new cell can reach them, then the new cell becomes usable.

Sorry if the wording is weird, but what I essentially mean is that you are making a redundant check with the part(line 83):

if (… && (grid[r][c] == opposite_direction[k] || grid[r][c] == ‘?’) { …

This was essential for the bfs because it was attempting to make the adjacent cells usable from a currently usable cell. In the reverse updates with the queries, you are trying to make the current cell usable from and adjacency usable cell, making this not a necessary requirement.

After editing your code by removing just that check and not changing anything else, your code AC’s(I tried submitting).

Hope that helps,
Eric Liu

1 Like
#3

Yeah that makes a lot of sense. The current cell is already a ‘?’ and can visit any of its neighbors, so ‘opposite_direction’ is not application in this case.

Thanks!

1 Like