HDU 4035：Maze（概率DP）

2021-07-01 23:12:30

http://acm.split.hdu.edu.cn/showproblem.php?pid=4035

Maze

Special Judge

Problem Description

When wake up, lxhgww find himself in a huge maze.

The maze consisted by N rooms and tunnels connecting these rooms. Each pair of rooms is connected by one and only one path. Initially, lxhgww is in room 1. Each room has a dangerous trap. When lxhgww step into a room, he has a possibility to be killed and restart from room 1. Every room also has a hidden exit. Each time lxhgww comes to a room, he has chance to find the exit and escape from this maze.

Unfortunately, lxhgww has no idea about the structure of the whole maze. Therefore, he just chooses a tunnel randomly each time. When he is in a room, he has the same possibility to choose any tunnel connecting that room (including the tunnel he used to come to that room).
What is the expect number of tunnels he go through before he find the exit?

Input

First line is an integer T (T ≤ 30), the number of test cases.

At the beginning of each case is an integer N (2 ≤ N ≤ 10000), indicates the number of rooms in this case.

Then N-1 pairs of integers X, Y (1 ≤ X, Y ≤ N, X ≠ Y) are given, indicate there is a tunnel between room X and room Y.

Finally, N pairs of integers Ki and Ei (0 ≤ Ki, Ei ≤ 100, Ki + Ei ≤ 100, K1 = E1 = 0) are given, indicate the percent of the possibility of been killed and exit in the ith room.

Output

For each test case, output one line “Case k: ”. k is the case id, then the expect number of tunnels lxhgww go through before he exit. The answer with relative error less than 0.0001 will get accepted. If it is not possible to escape from the maze, output “impossible”.

Sample Input

1 2

1 3

0 0

100 0

0 100

1 2

2 3

0 0

100 0

0 100

1 2

2 3

1 4

4 5

4 6

0 0

20 30

40 30

50 50

70 10

20 60

Sample Output

Case 1: 2.000000

Case 2: impossible

Case 3: 2.895522

题意：给出一棵树，从节点1走起，每个节点有一个被杀的几率k[i]，被杀的话就要回到节点1从头开始，还有一个逃脱几率e[i]，可以逃脱迷宫，与该节点相连的共有m条边，走任意一条边的几率相等（包括走回头路），求走出迷宫所走的边数的期望。

思路：这道题想了好久实在不会做，只能去看kuangbin菊苣的题解了，大概理解了怎么做，好像有点树型DP的思想。等学一点树型DP再回来自己做。

http://www.cnblogs.com/kuangbin/archive/2012/10/03/2711108.html

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <cstring>

 #include <string>

 #include <cmath>

 #include <queue>

 #include <vector>

 using namespace std;

 #define N 10010

 const double eps = 1e-;

 double k[N], e[N];

 double A[N], B[N], C[N];

 vector<int> G[N];

 bool dfs(int u, int pre)

 {

     int l = G[u].size();

     A[u] = k[u];

     B[u] = ( - k[u] - e[u]) / l;

     C[u] =  - k[u] - e[u];

     double tmp = ;

     for(int i = ; i < l; i++) {

         int v = G[u][i];

         if(v == pre) continue;

         if(!dfs(v, u)) return false;

         A[u] += ( - k[u] - e[u]) * A[v] / l;

         C[u] += ( - k[u] - e[u]) * C[v] / l;

         tmp += ( - k[u] - e[u]) * B[v] / l ;

     }

     if(fabs(tmp - ) < eps) return false;

     A[u] /= ( - tmp);

     B[u] /= ( - tmp);

     C[u] /= ( - tmp);

     return true;

 }

 int main()

 {

     int t, cas = ;

     scanf("%d", &t);

     while(t--) {

         memset(A, , sizeof(A));

         memset(B, , sizeof(B));

         memset(C, , sizeof(C));

         int n;

         scanf("%d", &n);

         for(int i = ; i <= n; i++)

             G[i].clear();

         for(int i = ; i < n; i++) {

             int u, v;

             scanf("%d%d", &u, &v);

             G[u].push_back(v);

             G[v].push_back(u);

         }

         for(int i = ; i <= n; i++) {

            scanf("%lf%lf", &k[i], &e[i]);

            k[i] /= ; e[i] /= ;

         }

         printf("Case %d: ", cas++);

         if(dfs(, -) && fabs(A[] - ) > eps) {

             printf("%.6f\n",  C[] / ( - A[]));

         } else {

             puts("impossible");

         }

     }

     return ;

 }

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Maze

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