原题传送门

我们珂以先考虑一条链的情况，设\(sum\)为所有\(w_i\)的总和，\(Sw_i\)表示\(\sum_{j=i}^nw_i\)

\[1 \rightarrow 2 \rightarrow 3 \rightarrow …… \rightarrow n\]

\[P(1\rightarrow n)=\prod_{i=1}^n(\frac{w_i}{Sum}\sum_{j=0}^{\inf}(\frac{Sum-Sw_i}{Sum})^j)=\prod_{i=1}^n\frac{w_i}{Sw_i}\]

考虑有反向边

\[1 \rightarrow 2 \rightarrow …… k \leftarrow k+1 \rightarrow …… \rightarrow n\]

由容斥原理可得：

\[P=P(1 \rightarrow k) \times P(k+1 \rightarrow n)-P(1 \rightarrow n)\]

我们珂以由链推广到树

设\(f[i][j]\)表示\(i\)的子树权值和为\(j\)时状态合法的概率

暴力dp即可

#include <bits/stdc++.h>
#define ll long long
#define N 1005
#define mod 998244353
#define getchar nc
using namespace std;
inline char nc(){static char buf[100000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{register int x=0,f=1;register char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x*f;
}
inline void write(register int x)
{if(!x)putchar('0');if(x<0)x=-x,putchar('-');static int sta[20];register int tot=0;while(x)sta[tot++]=x%10,x/=10;while(tot)putchar(sta[--tot]+48);
}
inline ll fastpow(register ll a,register int b)
{ll res=1;while(b){if(b&1)res=(res*a)%mod;b>>=1;a=(a*a)%mod;}return res;
}
struct edge{int to,next,st;
}e[N<<1];
int head[N],cnt;
inline void add(register int u,register int v,register int w)
{e[++cnt]=(edge){v,head[u],w};head[u]=cnt;
}
int n,p[N][4],inv[N*3],f[N][N*3],g[N*3],size[N],ans;
inline void dfs(register int x,register int fa)
{f[x][0]=1;for(register int i=head[x];i;i=e[i].next){int v=e[i].to;if(v==fa)continue;dfs(v,x);memset(g,0,sizeof(g));for(register int j=0;j<=size[x];++j)for(register int k=0;k<=size[v];++k){g[j+k]=(g[j+k]+1ll*f[x][j]*f[v][k]%mod*e[i].st%mod)%mod;if(e[i].st!=1)g[j]=(g[j]+1ll*f[x][j]*f[v][k]%mod)%mod;}memcpy(f[x],g,sizeof(g));size[x]+=size[v];}memset(g,0,sizeof(g));for(register int j=0;j<=size[x];++j)for(register int k=1;k<=3;++k)g[j+k]=(g[j+k]+1ll*f[x][j]*k%mod*inv[j+k]%mod*p[x][k]%mod)%mod;memcpy(f[x],g,sizeof(g));size[x]+=3;
}
int main()
{n=read();for(register int i=1;i<=n;++i){int x=0;for(register int j=1;j<=3;++j){p[i][j]=read();x+=p[i][j];}for(register int j=1;j<=3;++j)p[i][j]=p[i][j]*fastpow(x,mod-2)%mod;}for(register int i=1;i<n;++i){int u=read(),v=read();add(u,v,1);add(v,u,mod-1);}inv[1]=1;for(register int i=2;i<=n*3;++i)inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;dfs(1,0);for(register int i=1;i<=n*3;++i)ans=(ans+f[1][i])%mod;write(ans);return 0;
}