Problem: Example

Input

2 10 6
4 4 E
6 4 W


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

#define REP(i, n) for (int i=0; i<(n); ++i)
#define RREP(i, n) for (int i=(int)(n)-1; i>=0; --i)
#define FOR(i, a, n) for (int i=(a); i<(n); ++i)
#define RFOR(i, a, n) for (int i=(int)(n)-1; i>=(a); --i)

#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()

#define DUMP(x) cerr<<#x<<" = "<<(x)<<endl
#define DEBUG(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<<endl;

template<class T>
ostream &operator<<(ostream &os, const vector <T> &v) {
    os << "[";
    REP(i, SZ(v)) {
        if (i) os << ", ";
        os << v[i];
    }
    return os << "]";
}

template<class T, class U>
ostream &operator<<(ostream &os, const pair <T, U> &p) {
    return os << "(" << p.first << " " << p.second << ")";
}

template<class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

using ll = long long;
using ull = unsigned long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;

const ll MOD = 1e9 + 7;
const int INF = INT_MAX / 2;
const ll LINF = LLONG_MAX / 2;
const ld eps = 1e-9;

template<typename T>
struct edge {
    int src, to;
    T cost;
};

template<typename T>
using Graph = vector<vector<edge<T>>>;

using State = string::const_iterator;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    // ifstream in("in.txt");
    // cin.rdbuf(in.rdbuf());

    int n, w, d; cin >> n >> w >> d;
    vector<P> events;
    REP(i, n) {
        int x, y;
        char dir;
        cin >> x >> y >> dir;
        if (dir == 'W') {
            events.emplace_back(-x + y, x + y);
        }
        if (dir == 'N') {
            events.emplace_back(x + y, (d - y) - (w - x) + d + w);
        }
        if (dir == 'E') {
            events.emplace_back((d - y) - (w - x) + d + w, (d - y) + (w - x) + d + w);
        }
        if (dir == 'S') {
            events.emplace_back((d - y) + (w - x) + d + w, -x + y + 2 * d + 2 * w);
        }
    }
    sort(ALL(events), [&](P a, P b){ return a.second < b.second; });

    int ans = n;
    REP(i, SZ(events)) {
        auto tmp = events;
        REP(j, i) {
            if (tmp[j].second < tmp[i].second) {
                tmp[j].first += 2*d + 2*w;
                tmp[j].second += 2*d + 2*w;
            }
        }
        sort(ALL(tmp), [&](P a, P b){ return a.second < b.second; });
        int sum = 1, prev = tmp[0].second;
        FOR(k, 1, SZ(tmp)) {
            if (prev < tmp[k].first) {
                prev = tmp[k].second;
                ++sum;
            }
        }
        chmin(ans, sum);
    }
    cout << ans << endl;

    return 0;
}