結果

問題 No.2520 L1 Explosion
コンテスト
ユーザー karinohito
提出日時 2023-10-27 23:26:36
言語 C++17
(gcc 13.3.0 + boost 1.89.0)
結果
AC  
実行時間 110 ms / 2,000 ms
コード長 3,907 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 4,221 ms
コンパイル使用メモリ 268,576 KB
最終ジャッジ日時 2025-02-17 16:03:49
ジャッジサーバーID
（参考情報） 		judge1 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 22
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code 

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
#include <time.h> 
using namespace atcoder;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
#define all(A) A.begin(),A.end()
#define rep(i, n) for (ll i = 0; i < (ll) (n); i++)
template<class T>
bool chmax(T& p, T q, bool C = 1) {
    if (C == 0 && p == q) {
        return 1;
    }
    if (p < q) {
        p = q;
        return 1;
    }
    else {
        return 0;
    }
}
template<class T>
bool chmin(T& p, T q, bool C = 1) {
    if (C == 0 && p == q) {
        return 1;
    }
    if (p > q) {
        p = q;
        return 1;
    }
    else {
        return 0;
    }
}
ll modPow(long long a, long long n, long long p) {
    if (n == 0) return 1; // 0乗にも対応する場合
    if (n == 1) return a % p;
    if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;
    long long t = modPow(a, n / 2, p);
    return (t * t) % p;
}
ll cnt = 0;
ll gcd(ll(a), ll(b)) {
    cnt++;
    if (a == 0)return b;
    if (b == 0)return a;
    ll c = a;
    while (a % b != 0) {
        c = a % b;
        a = b;
        b = c;
    }
    return b;
}
ll sqrtz(ll N) {
    ll L = 0;
    ll R = sqrt(N) + 10000;
    while (abs(R - L) > 1) {
        ll mid = (R + L) / 2;
        if (mid * mid <= N)L = mid;
        else R = mid;
    }
    return L;
}


ll nzkon(ll N, ll K) {//
    return 0;
}


using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vm>;
using vvvm = vector<vvm>;


vector<mint> fact, factinv, inv;
const ll mod = 998244353;
void prenCkModp(ll n) {
    fact.resize(n + 5);
    factinv.resize(n + 5);
    inv.resize(n + 5);
    fact[0] = fact[1] = 1;
    factinv[0] = factinv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < n + 5; i++) {
        fact[i] = (fact[i - 1] * i);
        inv[i] = (mod - ((inv[mod % i] * (mod / i))));
        factinv[i] = (factinv[i - 1] * inv[i]);
    }
}
mint nCk(ll n, ll k) {
    if (n < k || k < 0) return 0;
    return (fact[n] * ((factinv[k] * factinv[n - k])));
}

bool DEB = 0;

vvm mul(vvm A, vvm B) {
    ll n = A.size();
    vvm res(n, vm(n, 0));
    rep(i, n)rep(j, n)rep(k, n)res[i][j] += A[i][k] * B[k][j];
    return res;
}


int main() {

    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    ll N, M;
    cin >> N >> M;
    vll L(N), R(N), U(N), D(N);
    map<ll, ll> YOKO, TATE;
    rep(i, N) {
        ll X, Y, H;
        cin >> X >> Y >> H;
        H = M - H;
        ll x = X + Y;
        ll y = X - Y;
        L[i] = x - H;
        R[i] = x + H;
        U[i] = y - H;
        D[i] = y + H;
        //cout << L[i] << " " << R[i] << " " << U[i] << " " << D[i] << endl;
        YOKO[L[i]] = YOKO[R[i]] = 1;
        TATE[U[i]] = TATE[D[i]] = 1;
    }
    vll PY, PT;
    ll yn = 0;
    for (auto y : YOKO) {
        YOKO[y.first] = yn;
        PY.push_back(y.first);
        yn++;
    }
    ll tn = 0;
    for (auto y : TATE) {
        TATE[y.first] = tn;
        PT.push_back(y.first);
        tn++;
    }
    vvll AR(yn + 3, vll(tn + 3, 0));
    rep(i, N) {
        L[i] = YOKO[L[i]];
        R[i] = YOKO[R[i]];
        U[i] = TATE[U[i]];
        D[i] = TATE[D[i]];
        AR[L[i]][U[i]]++;
        AR[R[i]][D[i]]++;
        AR[L[i]][D[i]]--;
        AR[R[i]][U[i]]--;
        //cout << L[i] << " " << R[i] << " " << U[i] << " " << D[i] << endl;
    }
    rep(i, yn + 2)rep(j, tn + 3)AR[i + 1][j] += AR[i][j];
    rep(i, yn + 3)rep(j, tn + 2)AR[i][j + 1] += AR[i][j ];

    vm AN(N + 1, 0);
    rep(i, yn - 1) {
        rep(j, tn - 1) {
            mint dy = (PY[i + 1] - PY[i]);
            mint dt= (PT[j + 1] - PT[j]);
            mint area = dy * dt;
            ll num = AR[i][j];
            AN[num] += area;
        }
    }

    rep(i, N) {
        AN[i + 1] /= mint(2);
        cout << AN[i + 1].val() << endl;
    }



}
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