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HDU 4035 Maze——概率dp


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。
思路:
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)
带入上面非叶子结点的式子得(这一步化简比较麻烦,这里直接给出了最终结果)
(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);

同理对于叶子结点:

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 <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
const int maxn = 1e4 + 10;
const double eps = 1e-10;
int T, n;
double k[maxn], e[maxn], a[maxn], b[maxn], c[maxn];
vector<int> G[maxn];
bool dfs(int u, int f) {
int m = G[u].size();
a[u] = k[u];
b[u] = (1-k[u]-e[u])/m;
c[u] = 1-k[u]-e[u];
double temp = 0;
for (int i = 0; i < m; i++) {
int v = G[u][i];
if (v == f) continue;
if (!dfs(v, u)) return false;
a[u] += (1-k[u]-e[u])/m*a[v];
c[u] += (1-k[u]-e[u])/m*c[v];
temp += (1-k[u]-e[u])/m*b[v];
}
if (fabs(1-temp) < eps) return false;
a[u] /= (1-temp);
b[u] /= (1-temp);
c[u] /= (1-temp);
return true;
}
int main() {
scanf("%d", &T);
for (int kase = 1; kase <= T; kase++) {
scanf("%d", &n);
for (int i = 0; i <= n; i++) G[i].clear();
int x, y;
for (int i = 1; i < n; i++) {
scanf("%d%d", &x, &y);
G[x].push_back(y);
G[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;
}
printf("Case %d: ", kase);
if (dfs(1, -1) && fabs(1-a[1]) > eps) printf("%.6f\n", c[1]/(1-a[1]));
else printf("impossible\n");
}
return 0;
}

 

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