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树形dp中的\text{up and down}

还有一点期望

设每个点能够通电的概率为P_i

那么ans = \sum_{i=1}^n P_i

我们先考虑up(只考虑子树))

分成两种情况:

  1. 直接通电 : q_i

  2. 被子节点通电 :

    (1-q_i) \times P_{to}(\text{子节点}) \times e_i.w(\text{连接的边导电的概率})

我们再考虑down(考虑子树外))

我们可以通过父节点的答案来更新子节点

父节点的答案是包括当前子节点的

可以把这部分除去然后按照up的方法更新

P_x= P_x' (1-P_x')*(P_{to}*e_i.w)(P_x'指根节点除去当前子节点的贡献的答案)

整理得: P_x'=\frac{P_x-(P_{to} \times e_i.w)}{1-(P_{to} \times e_i.w)}

(这里有个坑点:当1-(P_{to} \times e_i.w) = 0时这个式子就没有意义了)

\therefore P_{to} = P_{to} + (1-P_{to}) \times P_x' \times e_i.w

所以在下推的时候要特判一下

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 500011
int n,cnt=0,head[N],d[N];
struct edge{
    int to,nxt;
    db w;
}e[N*2];
il void add(int x,int y,db w){
    e[++cnt].to=y;e[cnt].w=w;e[cnt].nxt=head[x];head[x]=cnt;
}
db s[N],ans=0;
void up(int x){
    fl(i,x)if(!d[to]){
        d[to]=d[x]+1;
        up(to);
        db t=s[to]*e[i].w;
        s[x]+=(1-s[x])*t;
    }
}
#define eps 1e-7
il bool chk(db x){return x+eps>1.0&&x-eps<1.0;}
void down(int x){
    ans+=s[x];
    fl(i,x)if(d[to]>d[x]){
        if(!chk(s[to]*e[i].w)){
            db t=(s[x]-s[to]*e[i].w)/(1.0-s[to]*e[i].w)*e[i].w;
            s[to]+=(1-s[to])*t;
        }
        down(to);
    }
}
int main(){
    fin("in");
    in>>n;
    int x,y,w;
    Fur(i,1,n-1)in>>x>>y>>w,add(x,y,(db)w*0.01),add(y,x,(db)w*0.01);
    Fur(i,1,n)in>>x,s[i]=(db)x*0.01;
    d[1]=1;
    up(1);
    down(1);
    printf("%.6f\n",ans);
}
LG 4284 [SHOI2014]概率充电器
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