HDU 4035 Maze （概率DP）

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/*lxhgww被困在迷宫里，迷宫是一棵n顶点的树，lxhgww初始在点1。每个点三种可能    1.被杀，回到起点1（概率为k[i]）    2.逃脱，即逃出迷宫（概率为e[i]）    3.随机的走一条与改点相连的边（包括它与它父亲相连的那条边求逃出迷宫期望的步数。设E[i]表示在结点i时期望的步数, fi表示点i的父亲, m表示点i的度数，j表示i的子节点    当i为叶子结点时，E[i] = k[i] * E[1] + e[i] * 0 + (1-k[i]-e[i]) * (E[fi] + 1)                          = k[i] * E[1] + (1-k[i]-e[i]) * E[fi] + (1-k[i]-e[i])    当i为非叶子节点时, E[i] = k[i] * E[1] + e[i] * 0 + (1-k[i]-e[i]) * ( (E[fi]+∑E[j]) / m + 1 )                            = k[i] * E[1] + (1-k[i]-e[i])/m * E[fi] + (1-k[i]-e[i]) / m * ( ∑E[j] + m )令E[i] = A[i] * E[1] + B[i] * E[fi] + C[i]    当i为叶子结点时, A[i] = k[i], B[i] = (1-k[i]-e[i]), C[i] = (1-k[i]-e[i])    当i为非叶子结点时, E[i] = k[i] * E[1] + (1-k[i]-e[i]) * ( (E[fi] + ∑(A[j]*E[1] + B[j]*E[i] + C[j])) / m + 1 )                            = (k[i]+(1-k[i]-e[i])/m*∑A[j]) * E[1] + (1-k[i]-e[i])/m * E[fi] + ((1-k[i]-e[i])/m*∑B[j]) * E[i] + (∑C[j]+m)*(1-k[i]-e[i])/m;                       令(1-k[i]-e[i])/m*∑B[j] = X, AA = (k[i]+(1-k[i]-e[i])/m*∑A[j]), BB = (1-k[i]-e[i])/m, CC = (∑C[j]+m)*(1-k[i]-e[i])/m                       则E[i] = AA * E[1] + BB * E[fi] + X * E[i] + CC                              = AA/(1-X) * E[1] + BB/(1-X) * E[fi] + CC/(1-x)                       因此, A[i] = AA/(1-X), B[i] = BB/(1-X), C[i] = CC/(1-x)    逆推出A[1], B[1], C[1]，即可得到答案*/#include<cstdio>#include<cstring>#include<set>#include<algorithm>#include<vector>using namespace std;const int maxn = 1e4 + 10;const double eps = 1e-12;double A[maxn], B[maxn], C[maxn], dp[maxn];double K[maxn], E[maxn];vector<int> G[maxn];int n;void Cal(int u, int fa){double m = (1-K[u]-E[u]) / G[u].size();    if(G[u].size() == 1 && G[u][0] == fa) {    A[u] = K[u];    B[u] = (1-K[u]-E[u]);    C[u] = (1 - K[u] - E[u]);    return;    }    for(int i = 0; i < (int)G[u].size(); i++) {    int v = G[u][i];    if(v == fa) continue;    Cal(v, u);    }    double a = 0, b = 0, c = 0, x;    for(int i = 0; i < (int)G[u].size(); i++) {    int v = G[u][i];     if(v == fa) continue;        a += A[v]; b += B[v]; c += C[v];    }    b *= m;    x = 1 - b;    A[u] = (K[u] + m*a) / x;    B[u] = m / x;    C[u] = (c + G[u].size()) * m / x;}int main(){int T;scanf("%d", &T);for(int kase = 1; kase <= T; kase++){         scanf("%d", &n);         for(int i = 1; i <= n; i++) G[i].clear();         for(int i = 1; i <= n-1; i++) {         int u, v;         scanf("%d%d", &u, &v);         G[u].push_back(v);         G[v].push_back(u);         }         for(int i = 1; i <= n; i++) {         scanf("%lf%lf", &K[i], &E[i]);         K[i] /= 100.0;         E[i] /= 100.0;         }         Cal(1, 0);         printf("Case %d: ", kase);         if(1 - A[1] < eps) printf("impossible\n");         else printf("%.6f\n", C[1] / (1.0 - A[1]));}return 0;}
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