链接:http://acm.hdu.edu.cn/showproblem.php?pid=4035
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...
转载自博客:https://blog.csdn.net/morgan_xww/article/details/6776947/
代码:
#include <cstdio> #include <iostream> #include <vector> #include <cmath> using namespace std; const int MAXN = 10000 + 5; double e[MAXN], k[MAXN]; double A[MAXN], B[MAXN], C[MAXN]; vector<int> v[MAXN]; bool search(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 tmp = 0; for (int j = 0; j < (int)v[i].size(); j++) { if ( v[i][j] == fa ) continue; if ( !search(v[i][j], i) ) return false; A[i] += A[v[i][j]] * B[i]; C[i] += C[v[i][j]] * B[i]; tmp += B[v[i][j]] * B[i]; } if ( fabs(tmp - 1) < 1e-10 ) return false; A[i] /= 1 - tmp; B[i] /= 1 - tmp; C[i] /= 1 - tmp; return true; } int main() { int nc, n, s, t; cin >> nc; for (int ca = 1; ca <= nc; ca++) { cin >> n; for (int i = 1; i <= n; i++) v[i].clear(); for (int i = 1; i < n; i++) { cin >> s >> t; v[s].push_back(t); v[t].push_back(s); } for (int i = 1; i <= n; i++) { cin >> k[i] >> e[i]; k[i] /= 100.0; e[i] /= 100.0; } cout << "Case " << ca << ": "; if ( search(1, -1) && fabs(1 - A[1]) > 1e-10 ) cout << C[1]/(1 - A[1]) << endl; else cout << "impossible" << endl; } return 0; }