传送门

题意：

思路： 给你个$DAG$，由于每一条路径出现概率相等，那么期望就是$\frac{总长度}{路径个数}$。设$f[i]$表示到$i$这个点的总长度，$g[i]$表示到$i$这个点路径的总个数。那么转移方程也比较好想了：$$f[i]=\sum _{edge(j,i)}(f[j]+g[j])$$ $$g[i]=\sum _{edge(j,i)}g[j]+1$$
$g[i]$要加一是因为要加上自己到自己的，最终答案为$\frac{\sum f[i]}{\sum g[i]}$。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=998244353,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m;
int d[N];
LL f[N],g[N];
vector<int>v[N];LL qmi(LL a,LL b)
{LL ans=1;while(b){if(b&1) ans=ans*a%mod;a=a*a%mod;b>>=1;}return ans;
}void topsort()
{queue<int>q;for(int i=1;i<=n;i++) if(!d[i]) q.push(i),g[i]=1;while(q.size()){int u=q.front(); q.pop();for(auto x:v[u]){(f[x]+=f[u]+g[u])%=mod;(g[x]+=g[u])%=mod;if(--d[x]==0) q.push(x),g[x]++;}}LL ans1=0,ans2=0;for(int i=1;i<=n;i++) (ans1+=f[i])%=mod,(ans2+=g[i])%=mod;printf("%lld\n",ans1*qmi(ans2,mod-2)%mod);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){int a,b; scanf("%d%d",&a,&b);v[a].pb(b); d[b]++;}topsort();return 0;
}
/**/