[Cnoi2020]线性生物

期望入门题。但是我不会做。

考虑设\(E_{x\to{x+1}}\)为\(x\)到\(x+1\)点的期望步数。

则\(ans = \sum_{i = 0}^{n} E_{x\to{x+1}}\)

知\(E_{y\to{x+1}} = \sum_{i = y}^{x}E_{i\to{i + 1}}\)

\(E_{x\to{x+1}} = \frac{1}{son + 1} + \frac{1}{son + 1}\sum_{(x,y)\ in\ {Son}}(E_{y\to{x+1}} + 1)\)

设\(E_{x\to{x+1}} = f_x\),\(sum_x = \sum_i^xf_i\)

\(f_x = (son + 1) + \sum_{(x,y)\ in\ Son}sum_{x-1} - sum_{y - 1}\)

// Problem: P6835 [Cnoi2020]线形生物
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P6835
// Memory Limit: 128 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#define ll long long
#define mod 998244353
#define N 1000005

int head[N],cnt;
struct P{
	int to,next;
}e[N << 1];

inline void add(int x,int y){
	e[++cnt].to = y;
	e[cnt].next = head[x];
	head[x] = cnt;
}

ll n,m,k,out[N];
ll f[N],sum[N];

int main(){
	scanf("%lld%lld%lld",&k,&n,&m);
	for(int i = 1;i <= m;++i){
		ll l,r;
		scanf("%lld%lld",&l,&r);
		add(l,r);
		out[l]++;
	}
	for(int i = 1;i <= n;++i){
		f[i] = (out[i] + 1);
		for(int j = head[i];j;j = e[j].next){
			int v = e[j].to;
			f[i] = (f[i] + (sum[i - 1] - sum[v - 1]) % mod + mod) % mod;
		}
		sum[i] = (sum[i - 1] + f[i]) % mod;
	}
	std::cout<<sum[n] % mod<<std::endl;
	return 0;
}