码迷,mamicode.com
首页 > 其他好文 > 详细

hdu4035----Maze

时间:2014-12-22 09:34:31      阅读:152      评论:0      收藏:0      [点我收藏+]

标签:dp

Maze

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1788    Accepted Submission(s): 701
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
 

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: 405045132@qq.com 
    > 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;
}


hdu4035----Maze

标签:dp

原文地址:http://blog.csdn.net/guard_mine/article/details/42062973

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!