hdu4035_概率dp

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4035
dp求期望的题。
    题意：
    有n个房间，由n-1条隧道连通起来，实际上就形成了一棵树，
    从结点1出发，开始走，在每个结点i都有3种可能：
        1.被杀死，回到结点1处（概率为ki）
        2.找到出口，走出迷宫 （概率为ei）
        3.和该点相连有m条边，随机走一条
    求：走出迷宫所要走的边数的期望值。

    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[1]即为所求。

    叶子结点：
    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);

    非叶子结点：（m为与结点相连的边数）
    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);

    设对每个结点：E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;

    对于非叶子结点i，设j为i的孩子结点，则
    ∑(E[child[i]]) = ∑E[j]
                   = ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
                   = ∑(Aj*E[1] + Bj*E[i] + Cj)
    带入上面的式子得
    (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);

    对于叶子结点
    Ai = ki;
    Bi = 1 - ki - ei;
    Ci = 1 - ki - ei;

    从叶子结点开始，直到算出 A1,B1,C1;

    E[1] = A1*E[1] + B1*0 + C1;
    所以
    E[1] = C1 / (1 - A1);
    若 A1趋近于1则无解...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <vector>
using namespace std; 

const int N = 10000 + 10;
double e[N], k[N], a[N], b[N], c[N];
vector <int> v[N];
bool find(int i, int fa)
{
    if(v[i].size() == 1 && fa != -1)
    {
        a[i] = k[i];
        b[i] = 1 - k[i] - e[i];
        c[i] = 1 - k[i] - e[i];
        return true;
    }
    a[i] = k[i];
    b[i] = (1 - k[i] - e[i]) / v[i].size();
    c[i] = 1 - k[i] - e[i];
    double temp = 0;
    for(int j = 0; j < (int)v[i].size(); j++)
    {
        if(v[i][j] == fa)
            continue;
        if(!find(v[i][j], i))
            return false;
        a[i] += b[i] * a[v[i][j]];
        c[i] += b[i] * c[v[i][j]];
        temp += b[i] * b[v[i][j]];
    }
    if(fabs(1 - temp) < 1e-10)
        return false;
    a[i] /= (1 - temp);
    b[i] /= (1 - temp);
    c[i] /= (1 - temp);
    return true;
}
int main()
{
    int t, n, x, y, cnt = 0;
    scanf("%d", &t);
    while(t--)
    {
        cnt++;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            v[i].clear();
        for(int i = 1; i < n; i++)
        {
            scanf("%d %d", &x, &y);
            v[x].push_back(y);
            v[y].push_back(x);
        }
        for(int i = 1; i <= n; i++){
            scanf("%lf %lf", k + i, e + i);
            k[i] /= 100.0;
            e[i] /= 100.0;
        }
        cout << "Case " << cnt << ": "<<endl;
        if(find(1, -1) && 1 - a[1] > 1e-10)
            printf("%.6lf
", c[1]/(1 - a[1]));
        else
            cout << "impossible" << endl;
    }
    return 0;
}