hdu4035 Maze 题解

/*
    设 E[i]表示在结点i处，要走出迷宫所要走的边数的期望。
    E[i] = ki*E[1] + (1-ki-ei)*E[fa[i]] + (1-ki-ei);
    E[i] = ki*E[1] + (1-ki-ei)/siz[i]*E[fa[i]] + (1-ki-ei)/siz[i]*∑(E[child[i]]) + (1-ki-ei);
    设对每个结点：E[i] = Ai*E[1] + Bi*E[fa[i]] + Ci;
    ∑(E[child[i]])  = ∑E[j] = ∑(Aj*E[1] + Bj*E[i] + Cj)
    带入得
    (1 - (1-ki-ei)/siz[i]*∑Bj)*E[i] = (ki+(1-ki-ei)/siz[i]*∑Aj)*E[1] + (1-ki-ei)/siz[i]*E[fa[i]] + (1-ki-ei) + (1-ki-ei)/siz[i]*∑Cj;
    Ai = (ki+(1-ki-ei)/siz[i]*∑Aj) / (1 - (1-ki-ei)/siz[i]*∑Bj);
    Bi = (1-ki-ei)/siz[i] / (1 - (1-ki-ei)/siz[i]*∑Bj);
    Ci = ( (1-ki-ei)+(1-ki-ei)/siz[i]*∑Cj ) / (1 - (1-ki-ei)/siz[i]*∑Bj);
    E[1] = A1*E[1] + B1*0 + C1;
    E[1] = C1 / (1 - A1);
    上述式子中若分母为零则无解
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn=1e4+5;

struct edge
{
    int v,nxt;
}ep[maxn<<1];
int hd[maxn],kh;
double k[maxn],e[maxn],a[maxn],b[maxn],c[maxn];
int n,t,siz[maxn];
void add(int u,int v) {ep[++kh]=(edge){v,hd[u]};hd[u]=kh;}

bool dfs(int u,int fa)
{
    double x1=0;
    a[u]=k[u];b[u]=(1-k[u]-e[u])/siz[u];c[u]=1-k[u]-e[u];
    for(int i=hd[u];i!=-1;i=ep[i].nxt)
    {
        if(ep[i].v==fa) continue;
        if(!dfs(ep[i].v,u)) return 0;
        a[u]+=b[u]*a[ep[i].v];
        x1+=b[u]*b[ep[i].v];
        c[u]+=b[u]*c[ep[i].v];
    }
    if(1-x1<1e-9) return 0;
    a[u]/=1-x1;b[u]/=1-x1;c[u]/=1-x1;
    return 1;
}

int main()
{
    int u,v;
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        memset(hd,-1,sizeof(hd));kh=0;memset(siz,0,sizeof(siz));
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            add(u,v);add(v,u);siz[u]++;siz[v]++;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",k+i,e+i);
            k[i]/=100;e[i]/=100;
        }
        printf("Case %d: ",ca);
        if(dfs(1,0)&&fabs(1-a[1])>1e-9) printf("%lf\n",c[1]/(1-a[1]));
        else printf("impossible\n");
    }
    return 0;
}