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Difficulty

演算法難度4，思維難度5，程式碼難度4

Description

給定一棵樹，每個點有直接充電的概率，每條邊有傳遞充電的概率

求被充電的點的個數的期望值

$1\le n\le 5\times 10^5$

Solution

換根dp裸題咯

定義$dp_x$代表點$x$充不上點的概率，$ans=n-\sum dp_i$

轉移方程：$dp_x=(1-p_x)\times\prod_{u\in son(x)}(1-val_i+val_i\times dp_u)$

換根的式子直接從dp轉移方程就能看出來的

聽說卡精度？我沒什麼感覺。

時間複雜度$O(n)$

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<cstring>
#define LL long long
using namespace std;
inline int read()
 
{
    int x=0,f=1;char ch=' ';
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    return f==1?x:-x;
}
const int N=1e6+5;
int n,tot;
int head[N],to[N],Next[N];
double ans,val[N],p[N],dp[N];
inline
 
 void addedge(int x,int y,double c){
    to[++tot]=y;
    Next[tot]=head[x];
    head[x]=tot;
    val[tot]=c;
}
inline void dfs(int x,int fa){
    dp[x]=1.0-p[x];
    for(int i=head[x];i;i=Next[i]){
        int u=to[i];
        if(u==fa)continue;
        dfs(u,x);
        dp[x]*=1.0-val[i]+val[i]*dp[u];
    }
}
inline void dfs2(int x,int fa){
    ans+=1.0-dp[x];
    for(int i=head[x];i;i=Next[i]){
        int u=to[i];
        if(u==fa)continue;
        dp[x]/=1.0-val[i]+val[i]*dp[u];
        dp[u]*=1.0-val[i]+val[i]*dp[x];
        dfs2(u,x);
        dp[u]/=1.0-val[i]+val[i]*dp[x];
        dp[x]*=1.0-val[i]+val[i]*dp[u];
    }
}
int main(){
    n=read();
    for(int i=1;i<n;++i){
        int x=read(),y=read(),c=read();
        double l=(double)c/100.0;
        addedge(x,y,l);addedge(y,x,l);
    }
    for(int i=1;i<=n;++i){
        int c=read();
        p[i]=(double)c/100.0;
    }
    dfs(1,0);
    dfs2(1,0);
    printf("%.6lf\n",ans);
    return 0;
}