DMOJ · ccc26j4

CCC 2026 J4 Snail Path

This C++ solution uses data structures for CCC 2026 J4 Snail Path. Read the reasoning, inspect the code, or try your own test case below.

ccc26j4CCC 2026 J4MathData structuresC++45 lines
Solution194of 248
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Approach

Data structures

Snail Path matches the turn-distance instructions and the code's count of revisited grid positions.

Math

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

J_4_Snail_Path.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1LL << 60; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        set<pair<int, int>> s;
        int m; cin >> m;
        set<pair<int, int>> vis;
        pair<int, int> cur = {200000, 200000};
        vis.insert({cur.first, cur.second});
        int cnt = 0;
        for (int j = 0; j < m; j++) {
            char c; int x; cin >> c >> x;
            if (c == 'N') {
                for (int i = cur.second + 1; i <= cur.second + x; i++) {
                    if (vis.count(make_pair(cur.first, i))) cnt++;
                    vis.insert({cur.first, i});
                }
                cur = make_pair(cur.first, cur.second + x);
            } else if (c == 'S') {
                for (int i = cur.second - 1; i >= cur.second - x; i--) {
                    if (vis.count(make_pair(cur.first, i))) cnt++;
                    vis.insert({cur.first, i});
                }
                cur = make_pair(cur.first, cur.second - x);
            } else if (c == 'E') {
                for (int i = cur.first + 1; i <= cur.first + x; i++) {
                    if (vis.count(make_pair(i, cur.second))) cnt++;
                    vis.insert({i, cur.second});
                }
                cur = make_pair(cur.first + x, cur.second);
            } else {
                for (int i = cur.first - 1; i >= cur.first - x; i--) {
                    if (vis.count(make_pair(i, cur.second))) cnt++;
                    vis.insert({i, cur.second});
                }
                cur = make_pair(cur.first - x, cur.second);
            }
        }
        cout << cnt << '\n';
        return 0;
    }
        

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