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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#386245 | #8570. Idola-Tree | hos_lyric | TL | 1ms | 3800kb | C++14 | 9.6kb | 2024-04-11 14:31:46 | 2024-04-11 14:31:46 |
Judging History
answer
#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
using namespace std;
using Int = long long;
template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }
#define COLOR(s) ("\x1b[" s "m")
////////////////////////////////////////////////////////////////////////////////
template <unsigned M_> struct ModInt {
static constexpr unsigned M = M_;
unsigned x;
constexpr ModInt() : x(0U) {}
constexpr ModInt(unsigned x_) : x(x_ % M) {}
constexpr ModInt(unsigned long long x_) : x(x_ % M) {}
constexpr ModInt(int x_) : x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_) {}
constexpr ModInt(long long x_) : x(((x_ %= static_cast<long long>(M)) < 0) ? (x_ + static_cast<long long>(M)) : x_) {}
ModInt &operator+=(const ModInt &a) { x = ((x += a.x) >= M) ? (x - M) : x; return *this; }
ModInt &operator-=(const ModInt &a) { x = ((x -= a.x) >= M) ? (x + M) : x; return *this; }
ModInt &operator*=(const ModInt &a) { x = (static_cast<unsigned long long>(x) * a.x) % M; return *this; }
ModInt &operator/=(const ModInt &a) { return (*this *= a.inv()); }
ModInt pow(long long e) const {
if (e < 0) return inv().pow(-e);
ModInt a = *this, b = 1U; for (; e; e >>= 1) { if (e & 1) b *= a; a *= a; } return b;
}
ModInt inv() const {
unsigned a = M, b = x; int y = 0, z = 1;
for (; b; ) { const unsigned q = a / b; const unsigned c = a - q * b; a = b; b = c; const int w = y - static_cast<int>(q) * z; y = z; z = w; }
assert(a == 1U); return ModInt(y);
}
ModInt operator+() const { return *this; }
ModInt operator-() const { ModInt a; a.x = x ? (M - x) : 0U; return a; }
ModInt operator+(const ModInt &a) const { return (ModInt(*this) += a); }
ModInt operator-(const ModInt &a) const { return (ModInt(*this) -= a); }
ModInt operator*(const ModInt &a) const { return (ModInt(*this) *= a); }
ModInt operator/(const ModInt &a) const { return (ModInt(*this) /= a); }
template <class T> friend ModInt operator+(T a, const ModInt &b) { return (ModInt(a) += b); }
template <class T> friend ModInt operator-(T a, const ModInt &b) { return (ModInt(a) -= b); }
template <class T> friend ModInt operator*(T a, const ModInt &b) { return (ModInt(a) *= b); }
template <class T> friend ModInt operator/(T a, const ModInt &b) { return (ModInt(a) /= b); }
explicit operator bool() const { return x; }
bool operator==(const ModInt &a) const { return (x == a.x); }
bool operator!=(const ModInt &a) const { return (x != a.x); }
friend std::ostream &operator<<(std::ostream &os, const ModInt &a) { return os << a.x; }
};
////////////////////////////////////////////////////////////////////////////////
constexpr unsigned MO = 998244353;
using Mint = ModInt<MO>;
// sub[u]: inside subtree at u, rooted at u
// bus[u]: outside subtree at u, rooted at par[u]
// tot[u]: rooted at u
// Edge: edge information
// T: monoid representing information of a subtree.
// T::init() should assign the identity.
// T::pull(const T &l, const T &r) should assign the product.
// T::up(int u, int p, const Edge &e)
// should attach vertex u to the product of the subtrees,
// and if p != -1 then attach edge information e: u -> p.
template <class Edge, class T> struct ForestDPE {
int n;
vector<vector<pair<Edge, int>>> graph;
vector<int> par;
vector<T> sub, bus, tot;
ForestDPE() : n(0) {}
explicit ForestDPE(int n_) : n(n_), graph(n_) {}
void ae(int u, int v, const Edge &e0, const Edge &e1) {
assert(0 <= u); assert(u < n);
assert(0 <= v); assert(v < n);
graph[u].emplace_back(e0, v);
graph[v].emplace_back(e1, u);
}
void run() {
par.assign(n, -2);
sub.resize(n);
bus.resize(n);
tot.resize(n);
for (int u = 0; u < n; ++u) if (par[u] == -2) {
dfs0(u, -1);
dfs1(u, -1);
}
}
void dfs0(int u, int p) {
par[u] = p;
const int deg = graph[u].size();
int w = -1;
int jp = -1;
for (int j = deg; --j >= 0; ) {
const int v = graph[u][j].second;
if (p != v) {
dfs0(v, u);
if (~w) {
bus[v].pull(sub[v], bus[w]);
} else {
bus[v] = sub[v];
}
w = v;
} else {
jp = j;
}
}
if (~w) {
sub[u] = bus[w];
} else {
sub[u].init();
}
sub[u].up(u, p, (~jp) ? graph[u][jp].first : Edge());
}
void dfs1(int u, int p) {
const int deg = graph[u].size();
int v = -1, jv = -1;
for (int j = 0; j < deg; ++j) {
const int w = graph[u][j].second;
if (p != w) {
if (~v) {
bus[v].pull(tot[v], bus[w]);
bus[v].up(u, v, graph[u][jv].first);
tot[w].pull(tot[v], sub[v]);
dfs1(v, u);
} else {
if (~p) {
tot[w] = bus[u];
} else {
tot[w].init();
}
}
v = w; jv = j;
}
}
if (~v) {
bus[v] = tot[v];
bus[v].up(u, v, graph[u][jv].first);
tot[u].pull(tot[v], sub[v]);
dfs1(v, u);
} else {
if (~p) {
tot[u] = bus[u];
} else {
tot[u].init();
}
}
tot[u].up(u, -1, Edge());
}
};
////////////////////////////////////////////////////////////////////////////////
/*
leaf-edge: w[i]: X[i]-Y[i]
other: 1
contributions
1^2: const
w[i]^2: (N-1)
1 * 1: const
w[i] * 1: depends on i
w[i] * w[j]: 1
2 \sum[i<j] w[i] w[j] = (\sum[i] w[i])^2 - \sum[i] w[i]^2
*/
int N, C;
vector<int> A, B;
vector<int> deg;
struct Data {
Int s0, s1;
Mint s2;
// without leaf-edge
Int t0, t1, dt1;
Data() {}
void init() {
s0 = s1 = 0;
s2 = 0;
t0 = t1 = dt1 = 0;
}
void pull(const Data &a, const Data &b) {
assert(this != &a);
assert(this != &b);
s0 = a.s0 + b.s0;
s1 = a.s1 + b.s1;
s2 = a.s2 + b.s2;
t0 = a.t0 + b.t0;
t1 = a.t1 + b.t1;
dt1 = a.dt1 + b.dt1;
}
void up(int u, int p, int e) {
s2 += s1;
s1 += s0;
s0 += 1;
t1 += dt1;
t0 += 1;
dt1 = e * t0;
}
};
// merge (a[i], a[i] + d, a[i] + 2 d, ...)
template <class F> void mergeArithBrute(const vector<Int> &as, Int d, int n, F f) {
const int asLen = as.size();
priority_queue<Int, vector<Int>, greater<Int>> que;
for (int i = 0; i < asLen; ++i) que.push(as[i]);
for (int j = 0; j < n; ++j) {
const Int b = que.top();
que.pop();
f(b);
que.push(b + d);
}
}
int main() {
for (int numCases; ~scanf("%d", &numCases); ) { for (int caseId = 1; caseId <= numCases; ++caseId) {
scanf("%d%d", &N, &C);
A.resize(N - 1);
B.resize(N - 1);
for (int i = 0; i < N - 1; ++i) {
scanf("%d%d", &A[i], &B[i]);
--A[i];
--B[i];
}
Mint ans = 0;
if (N == 2) {
for (int c = N - 1; c <= C; ++c) {
const Mint now = (Int)c * c;
ans += now*now*now;
}
} else {
deg.assign(N, 0);
for (int i = 0; i < N - 1; ++i) {
++deg[A[i]];
++deg[B[i]];
}
int rt = -1;
vector<int> X;
for (int u = 0; u < N; ++u) {
if (deg[u] == 1) {
X.push_back(u);
} else {
if (!~rt) rt = u;
}
}
const int L = X.size();
ForestDPE<int, Data> f(N);
for (int i = 0; i < N - 1; ++i) {
const int e = (deg[A[i]] > 1 && deg[B[i]] > 1) ? 1 : 0;
f.ae(A[i], B[i], e, e);
}
f.run();
// for(int u=0;u<N;++u)cerr<<u<<": "<<f.tot[u].s0<<" "<<f.tot[u].s1<<" "<<f.tot[u].s2<<"; "<<f.tot[u].t0<<" "<<f.tot[u].t1<<endl;
// all 1
Mint base = 0;
for (int u = 0; u < N; ++u) {
base += f.tot[u].s1;
base += 2 * f.tot[u].s2;
}
base /= 2;
// (N-2) w^2 + p w
vector<Int> ps(L);
for (int i = 0; i < L; ++i) {
const int x = X[i];
const int y = f.graph[x][0].second;
ps[i] = 2 * f.tot[y].t1;
}
// cerr<<"base = "<<base<<", ps = "<<ps<<endl;
// diff: (N-2) (2w+1) + p
for (int i = 0; i < L; ++i) {
ps[i] += (N-2) * 3;
}
Mint now = base;
int c = L;
mergeArithBrute(ps, 2 * (N-2), C - (N-1) + 1, [&](Int t) -> void {
// cerr<<"c = "<<c<<": now = "<<now<<", t = "<<t<<endl;
ans += now*now*now;
now += t;
// c -> c+1
now += (c + c + 1);
++c;
});
}
printf("%u\n", ans.x);
}
#ifndef LOCAL
break;
#endif
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3800kb
input:
2 4 3 1 4 1 3 2 1 4 4 1 4 1 3 2 1
output:
3375 25327
result:
ok 2 tokens
Test #2:
score: 0
Accepted
time: 0ms
memory: 3756kb
input:
4 4 3 1 4 1 3 2 1 4 4 1 4 1 3 2 1 5 4 1 4 1 3 1 2 5 4 5 5 1 4 1 3 1 2 5 4
output:
3375 25327 54872 249984
result:
ok 4 tokens
Test #3:
score: -100
Time Limit Exceeded
input:
4 300000 50000000 216838 200677 44849 12926 125086 157230 26195 29767 241694 21336 21336 24457 105861 84565 184655 45583 175336 97945 286044 30927 295273 249694 109469 1566 193560 251229 176229 288707 206166 13532 166218 264413 299977 95587 159105 48116 57912 82606 97296 193689 115029 121192 68068 1...