[2022CCPC广州] Infection

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15.[2022CCPC广州] Infection02-20

16.[2022CCPC广州] XOR Sum02-20

一眼树形dp，但是因为有初始感染节点这个东西，所以可以添加一维，表示这个连通块是否确定了初始感染节点，也就是 $d p_{i, j, 0 / 1}$ 表示 $i$ 子树（选了 $i$ 节点）内选了 $j$ 个点进入连通块，是/否确定了初始感染节点的概率，初始值即为 $d p_{t, 1, 1} = \frac{a_{t}}{\sum a_{i}}, d p_{t, 1, 0} = p_{t}, d p_{t, 0, 0} = 1 - p_{t}$ ，然后树上背包转移就行了

code

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#pragma GCC optimeze(3)
#pragma GCC optimeze(2)
#define PII pair<int, int>
#define pb push_back
#define fi first
#define se second
#define lowbit(x) (x & (-x))
#define inv(x) (qpow(x,mod-2))
#define lwz lower_bound
#define blong(i) ((i+K-1)/K)
using namespace std;
const int N=2e3+5;
const int M=3e2+5;
const int mod=1e9+7;
double eps=1e-6;
inline int read(){
	char ch=getchar();bool f=0;int x=0;
	for(;!isdigit(ch);ch=getchar())if(ch=='-')f=1;
	for(;isdigit(ch);ch=getchar())x=(x<<1)+(x<<3)+(ch^48);
	if(f==1)x=-x;return x;
}
ll qpow(ll a,ll b){
	ll ans=1;
	while(b){
		if(b&1)ans*=a,ans%=mod;
		a*=a,a%=mod,b>>=1;
	}
	return ans;
}
int gcd(int a,int b){return b==0? a:gcd(b,a%b);}
int n,a[N],p[N],sum,dp[N][N][2],siz[N],ans[N],f[N][2];
vector<int>G[N];
void add(int&a,int b){a+=b;if(a>=mod)a-=mod;}
void minus(int&a,int b){a-=b;if(a<0)a+=mod;}
void dfs(int t,int fa){
	dp[t][1][1]=1ll*a[t]*qpow(sum,mod-2)%mod,dp[t][1][0]=p[t],dp[t][0][0]=(1-p[t]+mod)%mod;
	siz[t]=1;
	for(auto to:G[t]){
		if(to==fa)continue;
		dfs(to,t);
		for(int i=0;i<=siz[t]+siz[to];i++)f[i][0]=f[i][1]=0;
		for(int i=1;i<=siz[t];i++){
			for(int j=0;j<=siz[to];j++){
				add(f[i+j][0],1ll*dp[t][i][0]*dp[to][j][0]%mod);
				add(f[i+j][1],1ll*dp[t][i][1]*dp[to][j][0]%mod);
				add(f[i+j][1],1ll*dp[t][i][0]*dp[to][j][1]%mod);
			}
		}
		siz[t]+=siz[to];
		for(int i=1;i<=siz[t];i++)dp[t][i][0]=f[i][0],dp[t][i][1]=f[i][1];
	}
	for(int i=1;i<=siz[t];i++)add(ans[i],1ll*dp[t][i][1]*((1-p[fa])%mod+mod)%mod);
}
signed main(){ 
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n;
	for(int i=1;i<n;i++){
		int u,v;cin>>u>>v;
		G[u].pb(v),G[v].pb(u);
	}
	for(int i=1;i<=n;i++){
		int x,y;cin>>a[i]>>x>>y;
		p[i]=1ll*x*qpow(y,mod-2)%mod;add(sum,a[i]);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		cout<<ans[i]<<'\n';
	}
	return 0;
}