AtCoder Beginner Contest 213 H Stroll

洛谷传送门

AtCoder 传送门

考虑一个朴素 dp，\(f_{t, u}\) 表示 \(t\) 时刻走到 \(u\) 点的方案数。有转移：

\[f_{t, u} = \sum\limits_{(u, v) = E_i} \sum\limits_{k = 0}^{t - 1} f_{k, v} \times p_{i, t - k} \]

直接做时间复杂度 \(O(mT^2)\)，无法接受。

发现转移是加法卷积形式，又因为这个 dp 是在线的，考虑分治 NTT。设当前递归区间为 \([l, r]\)，设 \(mid = \left\lfloor\frac{l + r}{2}\right\rfloor\)，计算出 \(f_{l \sim mid, v}\) 后，卷上 \(p_{i, 0 \sim r - l}\) 可以转移至 \(f_{mid + 1 \sim r, u}\)。时间复杂度降至 \(O(mT \log^2 T)\)，可以通过。

code

// Problem: H - Stroll
// Contest: AtCoder - AtCoder Beginner Contest 213
// URL: https://atcoder.jp/contests/abc213/tasks/abc213_h
// Memory Limit: 1024 MB
// Time Limit: 5000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 200100;
const ll mod = 998244353;
const ll G = 3;

inline ll qpow(ll b, ll p) {
	ll res = 1;
	while (p) {
		if (p & 1) {
			res = res * b % mod;
		}
		b = b * b % mod;
		p >>= 1;
	}
	return res;
}

ll n, m, K, a[15][maxn], r[maxn], f[maxn][15];
vector<pii> E[maxn];

typedef vector<ll> poly;

inline poly NTT(poly a, int op) {
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		if (i < r[i]) {
			swap(a[i], a[r[i]]);
		}
	}
	for (int k = 1; k < n; k <<= 1) {
		ll wn = qpow(op == 1 ? G : qpow(G, mod - 2), (mod - 1) / (k << 1));
		for (int i = 0; i < n; i += (k << 1)) {
			ll w = 1;
			for (int j = 0; j < k; ++j, w = w * wn % mod) {
				ll x = a[i + j], y = w * a[i + j + k] % mod;
				a[i + j] = (x + y) % mod;
				a[i + j + k] = (x - y + mod) % mod;
			}
		}
	}
	if (op == -1) {
		ll inv = qpow(n, mod - 2);
		for (int i = 0; i < n; ++i) {
			a[i] = a[i] * inv % mod;
		}
	}
	return a;
}

inline poly operator * (poly a, poly b) {
	a = NTT(a, 1);
	b = NTT(b, 1);
	int n = (int)a.size();
	for (int i = 0; i < n; ++i) {
		a[i] = a[i] * b[i] % mod;
	}
	a = NTT(a, -1);
	return a;
}

void cdq(int l, int r) {
	if (l >= r) {
		return;
	}
	int mid = (l + r) >> 1, n = mid - l, m = r - l, k = 0;
	cdq(l, mid);
	while ((1 << k) <= n + m) {
		++k;
	}
	for (int i = 1; i < (1 << k); ++i) {
		::r[i] = (::r[i >> 1] >> 1) | ((i & 1) << (k - 1));
	}
	for (int u = 1; u <= ::n; ++u) {
		for (pii p : E[u]) {
			int v = p.fst, id = p.scd;
			poly A(1 << k), B(1 << k);
			for (int i = 0; i <= n; ++i) {
				A[i] = f[l + i][v];
			}
			for (int i = 0; i <= m; ++i) {
				B[i] = a[id][i];
			}
			poly C = A * B;
			for (int i = mid + 1; i <= r; ++i) {
				f[i][u] = (f[i][u] + C[i - l]) % mod;
			}
		}
	}
	cdq(mid + 1, r);
}

void solve() {
	scanf("%lld%lld%lld", &n, &m, &K);
	for (int i = 1, u, v; i <= m; ++i) {
		scanf("%d%d", &u, &v);
		E[u].pb(v, i);
		E[v].pb(u, i);
		for (int j = 1; j <= K; ++j) {
			scanf("%lld", &a[i][j]);
		}
	}
	f[0][1] = 1;
	cdq(0, K);
	printf("%lld\n", f[K][1]);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}