HDU4035-Maze

Maze

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65768/65768 K (Java/Others)

Problem Description

When wake up, lxhgww find himself in a huge maze.

The maze consisted by N rooms and tunnels connecting these rooms.

Each pair of rooms is connected by one and only one path. Initially, lxhgww is in room 1. Each room has a dangerous trap. When lxhgww step into a room, he has a possibility to be killed and restart from room 1. Every room also has a hidden exit. Each time lxhgww comes to a room, he has chance to find the exit and escape from this maze.

Unfortunately, lxhgww has no idea about the structure of the whole maze. Therefore, he just chooses a tunnel randomly each time. When he is in a room, he has the same possibility to choose any tunnel connecting that room (including the tunnel he used to come to that room).
What is the expect number of tunnels he go through before he find the exit?

Input

First line is an integer T (T ≤ 30), the number of test cases.

At the beginning of each case is an integer N (2 ≤ N ≤ 10000), indicates the number of rooms in this case.

Then N-1 pairs of integers X, Y (1 ≤ X, Y ≤ N, X ≠ Y) are given, indicate there is a tunnel between room X and room Y.

Finally, N pairs of integers Ki and Ei (0 ≤ Ki, Ei ≤ 100, Ki + Ei ≤ 100, K1 = E1 = 0) are given, indicate the percent of the possibility of been killed and exit in the ith room.

Output

For each test case, output one line “Case k: ”. k is the case id, then the expect number of tunnels lxhgww go through before he exit. The answer with relative error less than 0.0001 will get accepted. If it is not possible to escape from the maze, output “impossible”.

Sample Input

3
3
1 2
1 3
0 0
100 0
0 100
3
1 2
2 3
0 0
100 0
0 100
6
1 2
2 3
1 4
4 5
4 6
0 0
20 30
40 30
50 50
70 10
20 60

Sample Output

Case 1: 2.000000
Case 2: impossible
Case 3: 2.895522

题目大意

一棵树，根节点为1，人物(题面中所说的lxhgww)初始在根节点，每个节点有一个死亡概率和逃离概率。
人物死亡后传送回根节点，求人物成功逃离前期望走过边的条数。

思考

设E[i]为从i点出发，成功逃离前期望走过边的条数，k[i]为i点死亡概率，e[i]为i点逃离概率，t[i]为其他情况(即 t[i]=(1−k[i]−e[i]) ),Du[i]为i点的度,f[i]为i走到相邻点的概率(即 f[i]=t[i]Du[i] )

显然的，E[i]只与E[root],E[father],E[child]有关

E[leaf]=k[leaf]∗E[root]+e[leaf]∗0+t[i]∗(E[father]+1)

=k[leaf]∗E[root]+t[i]∗E[father]+t[i]

E[leaf]只与E[root],E[father]有关

那么叶子节点的父亲呢？叶子节点的父亲的父亲呢?……

通过子节点的与根和子节点的父亲(即当前节点)的关系，又可以得出当前节点与根和当前节点的父亲的关系

不妨设E[i]=A[i]∗E[root]+B[i]∗E[father]+C[i]

A[leaf]=k[leaf]

B[leaf]=C[leaf]=t[leaf]

E[i]=k[i]∗E[root]+e[i]∗0+f[i]∗(E[father]+1+∑(E[Child]+1))

E[i]=k[i]∗E[root]+f[i]∗E[father]+f[i]∗∑E[Child]+t[i]

E[i]=(k[i]+f[i]∗∑A[Child])∗E[root]+f[i]∗E[father]+(∑B[Child])∗E[i]+(t[i]+f[i]∗∑C[child])

]

(1−∑B[Child])∗E[i]=(k[i]+f[i]∗∑A[Child])∗E[root]+f[i]∗E[father]+(t[i]+f[i]∗∑C[child])

A[i]=k[i]+f[i]∗∑A[Child]1−∑B[Child]

B[i]=f[i]1−∑B[Child]

C[i]=t[i]+f[i]∗∑C[child]1−∑B[Child]

Ans=C[root]1−A[root]

无解情况即除数为0

代码

#include <cmath>#include <algorithm>#include <cstring>using namespace std;typedef long long ll;char ch;bool fl;inline void read(int &a){    for(fl=0,ch=getchar();ch<'0'||ch>'9';ch=getchar()) fl|=(ch=='-');    for(a=0;ch>='0'&&ch<='9';ch=getchar())a=(a<<3)+(a<<1)+(ch^'0');    if(fl)a=-a;}const int N=1e4+10;const double eps=1e-9;struct Edge{    int nd,nx;}e[N*2];double A[N],B[N],C[N];int k[N],p[N],Du[N],fir[N];int n,m,Case,x,y,tot=0;inline void addedge(int x,int y){e[++tot]=(Edge){y,fir[x]};fir[x]=tot;}double Abs(const double&a){ return a>0?a:-a;}bool Dfs(int x,int fa){    double temp=0,f;    int Now;    A[x]=0.01*k[x];C[x]=1-0.01*(k[x]+p[x]);    B[x]=f=C[x]/Du[x];    for(int i=fir[x];i;i=e[i].nx)    if(e[i].nd!=fa){        Now=e[i].nd;        if(!Dfs(Now,x)) return 0;        A[x]+=f*A[Now];C[x]+=f*C[Now];temp+=f*B[Now];    }    if(Abs(1-temp)<eps)return 0;    A[x]/=(1-temp);B[x]/=(1-temp);C[x]/=(1-temp);    return 1;} int main(){    read(Case);    for(int T=1;T<=Case;T++){        read(n);tot=0;        for(int i=1;i<=n;i++)fir[i]=0,Du[i]=0;        for(int i=1;i<n;i++){            read(x);read(y);            addedge(x,y);addedge(y,x);            Du[x]++;Du[y]++;        }         for(int i=1;i<=n;i++) read(k[i]),read(p[i]);        printf("Case %d: ",T);        if(Dfs(1,0)&&Abs(1-A[1])>eps){            printf("%lf\n",C[1]/(1-A[1]));        }        else printf("impossible\n");    }    return 0;}