HDU 4035 Maze（概率DP）

Maze

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 957    Accepted Submission(s): 307
Special Judge

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

 

 

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

 

 

Recommend

lcy

 

 

非常好的概率DP的題目。

詳細解釋見代碼的注釋中。

概率DP求期望的題目，主要是要列出方程，然后還要會對方程進行化簡，遞推。

/*
HDU 4035

    dp求期望的題。
    題意：
    有n個房間，由n-1條隧道連通起來，實際上就形成了一棵樹，
    從結點1出發，開始走，在每個結點i都有3種可能：
        1.被殺死，回到結點1處（概率為ki）
        2.找到出口，走出迷宮 （概率為ei）
        3.和該點相連有m條邊，隨機走一條
    求：走出迷宮所要走的邊數的期望值。

    設 E[i]表示在結點i處，要走出迷宮所要走的邊數的期望。E[1]即為所求。

    葉子結點：
    E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
         = ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei);

    非葉子結點：（m為與結點相連的邊數）
    E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
         = ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei);

    設對每個結點：E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;

    對於非葉子結點i，設j為i的孩子結點，則
    ∑(E[child[i]]) = ∑E[j]
                   = ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
                   = ∑(Aj*E[1] + Bj*E[i] + Cj)
    帶入上面的式子得
    (1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
    由此可得
    Ai =        (ki+(1-ki-ei)/m*∑Aj)   / (1 - (1-ki-ei)/m*∑Bj);
    Bi =        (1-ki-ei)/m            / (1 - (1-ki-ei)/m*∑Bj);
    Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);

    對於葉子結點
    Ai = ki;
    Bi = 1 - ki - ei;
    Ci = 1 - ki - ei;

    從葉子結點開始，直到算出 A1,B1,C1;

    E[1] = A1*E[1] + B1*0 + C1;
    所以
    E[1] = C1 / (1 - A1);
    若 A1趨近於1則無解...

*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<vector>
using namespace std;
const int MAXN=10010;
const double eps=1e-9;//這里1e-8會WA。設為1e-9和1e-10可以
double k[MAXN],e[MAXN];
double A[MAXN],B[MAXN],C[MAXN];

vector<int>vec[MAXN];//存樹

bool dfs(int t,int pre)//t的根結點是pre
{
    int m=vec[t].size();//點t的度
    A[t]=k[t];
    B[t]=(1-k[t]-e[t])/m;
    C[t]=1-k[t]-e[t];
    double tmp=0;
    for(int i=0;i<m;i++)
    {
        int v=vec[t][i];
        if(v==pre)continue;
        if(!dfs(v,t))return false;
        A[t]+=(1-k[t]-e[t])/m*A[v];
        C[t]+=(1-k[t]-e[t])/m*C[v];
        tmp+=(1-k[t]-e[t])/m*B[v];
    }
    if(fabs(tmp-1)<eps)return false;
    A[t]/=(1-tmp);
    B[t]/=(1-tmp);
    C[t]/=(1-tmp);
    return true;
}
int main()
{
   // freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    int T;
    int n;
    int u,v;
    int iCase=0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)vec[i].clear();
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            vec[u].push_back(v);
            vec[v].push_back(u);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&k[i],&e[i]);
            k[i]/=100;
            e[i]/=100;
        }
        printf("Case %d: ",iCase);
        if(dfs(1,-1)&&fabs(1-A[1])>eps)
        {
            printf("%.6lf\n",C[1]/(1-A[1]));
        }
        else printf("impossible\n");
    }
}