Maze

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

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

3
3
1 2
1 3
0 0
100 0
0 100
3
1 2
2 3
0 0
100 0
0 100
6
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

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online
Contest

Recommend

概率dp,状态方程想到了,但是没有去化简,看了kuangbin的博客做的,自己在纸上也试着推了一遍,看来以后做概率dp的时候如果感觉想到方程但没什么思路就多化化简.

/*************************************************************************
    > File Name: hdu4035.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2014年12月21日 星期日 17时19分33秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 10010;
const double eps = 1e-9;
int  n;
double k[N], e[N];
double A[N], B[N], C[N];
vector <int> edge[N];

bool dfs(int u, int fa)
{
	int m = edge[u].size();
	A[u] = m *  k[u];
	B[u] = (1 - k[u] - e[u]);
	C[u] = m * (1 - k[u] - e[u]);
	double f = m;
	for (int i = 0; i < m; ++i)
	{
		int v = edge[u][i];
		if (v == fa)
		{
			continue;
		}
		if (!dfs(v, u))
		{
			return false;
		}
		A[u] += (1 - k[u] - e[u]) * A[v];
		C[u] += (1 - k[u] - e[u])  * C[v];
		f -= (1 - k[u] - e[u]) * B[v];
	}
	if (abs(f) <= eps)
	{
		return false;
	}
	A[u] /= f;
	B[u] /= f;
	C[u] /= f;
	return true;
}

int main()
{
	int t;
	int u, v;
	scanf("%d", &t);
	int icase = 1;
	while (t--)
	{
		memset (A, 0, sizeof(A));
		memset (B, 0, sizeof(B));
		memset (C, 0, sizeof(C));
		scanf("%d", &n);
		for (int i = 1; i <= n; ++i)
		{
			edge[i].clear();
		}
		for (int i = 1; i <= n - 1; ++i)
		{
			scanf("%d%d", &u, &v);
			edge[u].push_back(v);
			edge[v].push_back(u);
		}
		for (int i = 1; i <= n; ++i)
		{
			scanf("%lf%lf", &k[i], &e[i]);
			k[i] /= 100;
			e[i] /= 100;
		}
		printf("Case %d: ", icase++);
		if (!dfs(1, -1) || abs(A[1] - 1) <= eps)
		{
			printf("impossible\n");
		}
		else
		{
			printf("%.6f\n", C[1] / (1 - A[1]));
		}
	}
	return 0;
}