NOI2021 路径交点

发布时间 2023-10-27 08:30:02作者: zltzlt

洛谷传送门

LOJ 传送门

两条路径的交点数量只和起点数量有关。容易发现是终点排列的逆序对数的奇偶性。求一个 \(f_{i, j}\) 表示从第 \(1\) 层的第 \(i\) 个点到第 \(k\) 层的第 \(j\) 个点的路径数量,对这个矩阵求行列式即可。

对于相交的路径数不用考虑,因为总存在和它对应的一条路径,满足两条路径的终点排列的逆序对数一奇一偶(LGV 引理的本质)。

code
// Problem: P7736 [NOI2021] 路径交点
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P7736
// Memory Limit: 1 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

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

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

const int maxn = 210;
const ll mod = 998244353;

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, a[maxn], b[maxn];

struct mat {
	ll n, m, a[maxn][maxn];
	mat(ll _n = 0, ll _m = 0) {
		n = _n;
		m = _m;
		mems(a, 0);
	}
} A;

inline mat operator * (const mat &a, const mat &b) {
	mat res(a.n, b.m);
	for (int k = 1; k <= a.m; ++k) {
		for (int i = 1; i <= a.n; ++i) {
			for (int j = 1; j <= b.m; ++j) {
				res.a[i][j] = (res.a[i][j] + a.a[i][k] * b.a[k][j]) % mod;
			}
		}
	}
	return res;
}

inline ll det(mat A) {
	ll ans = 1, n = A.n;
	for (int i = 1; i <= n; ++i) {
		for (int j = i; j <= n; ++j) {
			if (A.a[j][i]) {
				if (i != j) {
					swap(A.a[i], A.a[j]);
					ans = (mod - ans) % mod;
				}
				break;
			}
		}
		if (!A.a[i][i]) {
			return 0;
		}
		ans = ans * A.a[i][i] % mod;
		ll t = qpow(A.a[i][i], mod - 2);
		for (int j = i; j <= n; ++j) {
			A.a[i][j] = A.a[i][j] * t % mod;
		}
		for (int j = 1; j <= n; ++j) {
			if (i == j) {
				continue;
			}
			ll t = (mod - A.a[j][i]) % mod;
			for (int k = i + 1; k <= n; ++k) {
				A.a[j][k] = (A.a[j][k] + t * A.a[i][k]) % mod;
			}
		}
	}
	return ans;
}

void solve() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
	}
	for (int i = 1; i < n; ++i) {
		scanf("%lld", &b[i]);
	}
	A = mat(a[1], a[1]);
	for (int i = 1; i <= a[1]; ++i) {
		A.a[i][i] = 1;
	}
	for (int i = 1; i < n; ++i) {
		mat B(a[i], a[i + 1]);
		for (int _ = 0, u, v; _ < b[i]; ++_) {
			scanf("%d%d", &u, &v);
			B.a[u][v] = 1;
		}
		A = A * B;
	}
	printf("%lld\n", det(A));
}

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