题解：P15539 [CCC 2026 J4] Snail Path

模拟题。

蜗牛爬过的方格总数不超过 \(M\times 20+1\le 4\times 10^6+1\)。我们对无限网格建系，设蜗牛初始位置是 \((0,0)\)，上、下、左、右分别对应 \(y\) 轴正方向、\(y\) 轴负方向、\(x\) 轴负方向、\(x\) 轴正方向，使用哈希表（set）记录黏滑的方格，每次移动判断并进行统计即可。

时间复杂度 \(O(M\log M)\)（这里认为每次移动最多 \(20\) 步是常数）。

代码：

// Problem: P15539 [CCC 2026 J4] Snail Path
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P15539
// Memory Limit: 512 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {uniform_int_distribution<int> dist(L, R);return dist(rnd);
}template<typename T> void chkmin(T& x, T y) {if(y < x) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}template<int mod>
inline unsigned int down(unsigned int x) {return x >= mod ? x - mod : x;
}template<int mod>
struct Modint {unsigned int x;Modint() = default;Modint(unsigned int x) : x(x) {}friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}friend Modint operator/(Modint a, Modint b) {return a * ~b;}friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}friend Modint operator~(Modint a) {return a ^ (mod - 2);}friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}friend Modint& operator++(Modint& a) {return a += 1;}friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}friend Modint& operator--(Modint& a) {return a -= 1;}friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}friend bool operator==(Modint a, Modint b) {return a.x == b.x;}friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};set<tuple<int, int>> S;
int n, x, y, cnt;int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);S.insert({x, y});for(cin >> n; n; --n) {char dir; int steps;cin >> dir >> steps;if(dir == 'N') {while(steps--) {++y;if(S.count({x, y})) ++cnt;S.insert({x, y});}}else if(dir == 'S') {while(steps--) {--y;if(S.count({x, y})) ++cnt;S.insert({x, y});}}else if(dir == 'W') {while(steps--) {--x;if(S.count({x, y})) ++cnt;S.insert({x, y});}}else {while(steps--) {++x;if(S.count({x, y})) ++cnt;S.insert({x, y});}}}cout << cnt << endl;return 0;
}