HDU-4035 Maze (probabilistic DP seeking expectation)

The main topic: in a tree-shaped maze, the room as a node. There are n rooms, each room has the probability of a trap for KI, there is an exit probability of EI, if both cases are not present (the probability of pi), then can only make a choice to move to the next room (including may go to the previous room). The root node is 1, and when a trap is encountered, it must return to the root node 1 and start again, and when it encounters an exit, go out of the maze. Ask for the desired number of choices made from the beginning to the maze.

Topic Analysis: Define the State DP (i) to indicate the desired number of choices at node I until out of the maze. The state transition equation is:

DP (i) =KI*DP (1) + (1/m) *pi*∑ (DP (son) +1) (I is leaf node) <1>

DP (i) =KI*DP (1) + (1/m) *pi* (DP (father) +1) + (1/m) *pi* ∑ (DP (son) +1) (I is a non-leaf node) <2>

Will <2> tidy up, get:

DP (i) =KI*DP (1) + (1/m) *PI*DP (father) + (1/m) *pi*∑dp (son) +pi

Obviously, DP (i) is related to DP (1), and DP (i) =a (i) *DP (1) +b (i) *DP (father) +c (i) <3>

Bring <3> into ∑DP (son) to get:

(pi/m) *∑a (son)) dp (i) = (ki+ (pi/m) ∑a (son)) *DP (1) + (pi/m) *DP (father) + (pi/m) *∑c (son) +pi

Obviously, A (i), B (i), C (i) are available.

The DP (1) =C (1)/(1-a (1)) is the answer.

The code is as follows:

# include<iostream># include<cstdio># include<cmath># include<vector># include<cstring ># include<algorithm>using namespace Std;const int n=10005;const int inf=100000;const double eps=1e-9;int N;    Double K[n];d ouble e[n];d ouble a[n],b[n],c[n];vector<int>g[n];bool dfs (int u,int fa) {int m=g[u].size ();    Double temp=0;    A[u]=k[u];    B[u]= (1-k[u]-e[u])/m;    C[u]=1-k[u]-e[u];        for (int i=0;i<m;++i) {int v=g[u][i];        if (V==FA) continue;        if (!dfs (V,u)) return false;        a[u]+= (1-k[u]-e[u]) *a[v]/m;        c[u]+= (1-k[u]-e[u]) *c[v]/m;    temp+= (1-k[u]-e[u]) *b[v]/m;    } if (Fabs (1-temp) <eps) return false;    A[u]/= (1-temp);    B[u]/= (1-temp);    C[u]/= (1-temp); return true;}    int main () {int t,x,y,cas=0;    scanf ("%d", &t);        while (t--) {scanf ("%d", &n);        for (int i=1;i<=n;++i) g[i].clear ();            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;        e[i]/=100;        } printf ("Case%d:", ++cas);        if (Dfs (1,-1) &&fabs (1-a[1]) >eps) printf ("%.6lf\n", c[1]/(1-a[1]));    else printf ("impossible\n"); } return 0;}

　　

HDU-4035 Maze (probabilistic DP seeking expectation)