AtCoder Beginner Contest 242 Ex Random Painting

洛谷传送门

AtCoder 传送门

好久没复习过 min-max 容斥了……

设 \(i\) 被覆盖的时间为 \(t_i\)，考虑 min-max 容斥，那么：

\[E(\max\limits_{i \in S} t_i) = \sum\limits_{T \subseteq S \land T \ne \varnothing} (-1)^{|T| - 1} E(\min\limits_{i \in T} t_i) \]

\(\min\) 是好做的。设有 \(x\) 条线段与 \(T\) 有交，那么 \(E(\min\limits_{i \in T} t_i) = \frac{m}{x}\)。

考虑直接 dp，并且把 \(x\) 塞进状态里。设 \(f_{i, j}\) 表示考虑了 \([1, i]\)，钦定 \(i\) 被选，已经有 \(j\) 条线段与 \(T\) 无交的方案数，把容斥系数 \((-1)^{|T| - 1}\) 塞进转移里即可。

时间复杂度 \(O(n^2m)\)。

code

// Problem: Ex - Random Painting
// Contest: AtCoder - AtCoder Beginner Contest 242
// URL: https://atcoder.jp/contests/abc242/tasks/abc242_h
// Memory Limit: 1024 MB
// Time Limit: 3000 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 = 410;
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, m, f[maxn][maxn], g[maxn][maxn], h[maxn];
struct node {
	ll l, r;
} a[maxn];

void solve() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= m; ++i) {
		scanf("%lld%lld", &a[i].l, &a[i].r);
		++g[a[i].l][a[i].r];
	}
	for (int i = n; i; --i) {
		for (int j = 1; j <= n; ++j) {
			g[i][j] += g[i + 1][j] + g[i][j - 1] - g[i + 1][j - 1];
		}
	}
	f[0][0] = mod - 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= m; ++j) {
			for (int k = 0; k < i; ++k) {
				if (j >= g[k + 1][i - 1]) {
					f[i][j] = (f[i][j] - f[k][j - g[k + 1][i - 1]] + mod) % mod;
				}
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= m; ++j) {
			if (j >= g[i + 1][n]) {
				h[m - j] = (h[m - j] + f[i][j - g[i + 1][n]]) % mod;
			}
		}
	}
	ll ans = 0;
	for (int i = 1; i <= m; ++i) {
		ans = (ans + m * qpow(i, mod - 2) % mod * h[i] % mod) % mod;
	}
	printf("%lld\n", ans);
}

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