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

树上的概率dp。

 

Maze

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submission(s): 1626    Accepted Submission(s): 608 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
 

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

牛人的博客思路

dp求期望的题。
题意:
有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树,
从结点1出发,开始走,在每个结点i都有3种可能:
1.被杀死,回到结点1处(概率为ki)
2.找到出口,走出迷宫 (概率为ei)
3.和该点相连有m条边,随机走一条
求:走出迷宫所要走的边数的期望值。 设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望。E[1]即为所求。 叶子结点:
E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
= ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei); 非叶子结点:(m为与结点相连的边数)
E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
= ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei); 设对每个结点:E[i] = Ai*E[1] + Bi*E[father[i]] + Ci; 对于非叶子结点i,设j为i的孩子结点,则
∑(E[child[i]]) = ∑E[j]
= ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
= ∑(Aj*E[1] + Bj*E[i] + Cj)
带入上面的式子得
(1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
由此可得
Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj); 对于叶子结点
Ai = ki;
Bi = 1 - ki - ei;
Ci = 1 - ki - ei; 从叶子结点开始,直到算出 A1,B1,C1; E[1] = A1*E[1] + B1*0 + C1;
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int len,head[];
double A,B,C;
double k[],e[];
struct node
{
int now,next;
}tree[];
void add(int x,int y)
{
tree[len].now=y;
tree[len].next=head[x];
head[x]=len++;
}
void dfs(int root,int p)
{
int i,son,m=;
double a=,b=,c=,q;
for(i=head[root];i!=-;i=tree[i].next)
{
son=tree[i].now;
if(son==p)
{
continue;
}
dfs(son,root);
a+=A;
b+=B;
c+=C;
m++; } if(p != -)++m;
q=(-k[root]-e[root])/m;
A=(k[root]+q*a)/(-q*b);
B=q/(-q*b);
C=(-k[root]-e[root]+q*c)/(-q*b);
}
int main()
{
int t,n,a,b,j,i;
int x,y;
scanf("%d",&t);
for(j=;j<=t;j++)
{ len=;
memset(head,-,sizeof(head));
memset(e,,sizeof(e));
memset(k,,sizeof(k));
scanf("%d",&n);
for(i=;i<n;i++)
{
scanf("%d%d",&a,&b);
add(a,b);
add(b,a);
}
for(i=;i<=n;i++)
{
scanf("%d%d",&x,&y);
// printf("x=%d,y=%d\n",x,y);
k[i]=x/100.0;
e[i]=y/100.0;
// printf("k[i]=%lf,e[i]=%lf\n",k[i],e[i]);
}
dfs(,-);
if(-A<1e-)
printf("Case %d: impossible\n",j);
else
printf("Case %d: %lf\n",j,C/(-A));
}
return ;
}

HDU-4035 Maze的更多相关文章

  1. poj 2096 Collecting Bugs && ZOJ 3329 One Person Game && hdu 4035 Maze——期望DP

    poj 2096 题目:http://poj.org/problem?id=2096 f[ i ][ j ] 表示收集了 i 个 n 的那个. j 个 s 的那个的期望步数. #include< ...

  2. HDU 4035 Maze 概率dp,树形dp 难度:2

    http://acm.hdu.edu.cn/showproblem.php?pid=4035 求步数期望,设E[i]为在编号为i的节点时还需要走的步数,father为dfs树中该节点的父节点,son为 ...

  3. hdu 4035 Maze 概率DP

        题意:    有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树,    从结点1出发,开始走,在每个结点i都有3种可能:        1.被杀死,回到结点1处(概率为ki)      ...

  4. HDU 4035 Maze(树形概率DP)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4035 题意:一棵树,从结点1出发,在每个结点 i 都有3种可能:(1)回到结点1 , 概率 Ki:(2 ...

  5. hdu 4035 Maze(期待更多经典的树DP)

    Maze Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others) Total Submi ...

  6. HDU.4035.Maze(期望DP)

    题目链接 (直接)设\(F(i)\)为在\(i\)点走出迷宫的期望步数.答案就是\(F(1)\). 令\(p_i=1-k_i-e_i\),表示\(i\)点沿着边走的概率:\(d_i=dgr[i]\), ...

  7. HDU 4035 Maze 概率DP 搜索

    解题报告链接: http://www.cnblogs.com/kuangbin/archive/2012/10/03/2711108.html 先推公式,设计状态,令DP[i]表示在房间i退出要走步数 ...

  8. HDU 4035:Maze(概率DP)

    http://acm.split.hdu.edu.cn/showproblem.php?pid=4035 Maze Special Judge Problem Description   When w ...

  9. hdu 4035 2011成都赛区网络赛E 概率dp ****

    太吊了,反正我不会 /* HDU 4035 dp求期望的题. 题意: 有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树, 从结点1出发,开始走,在每个结点i都有3种可能: 1.被杀死,回到结点 ...

  10. hdu 5094 Maze 状态压缩dp+广搜

    作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4092176.html 题目链接:hdu 5094 Maze 状态压缩dp+广搜 使用广度优先 ...

随机推荐

  1. REST_FRAMEWORK加深记忆-加了用户登陆认证,自定义权限的API接口

    哈哈,终于快结束了.. urls.py from django.conf.urls import include, url from django.contrib import admin urlpa ...

  2. leetcode 4 : Median of Two Sorted Arrays 找出两个数组的中位数

    题目: There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the ...

  3. 关于Linux的时间与时区

    转:http://linux.chinaunix.net/techdoc/beginner/2007/06/22/960790.shtml 首先要说明的是我的系统是fedora,其他系统可能不完全相同 ...

  4. (转)java性能调优

    本文转自:http://blog.csdn.net/lilu_leo/article/details/8115612 一.类和对象使用技巧 1.尽量少用new生成新对象 用new创建类的实例时,构造雨 ...

  5. intellij idea 注释行如何自动缩进?

    我们指代代码注释的快捷键是 Ctrl+/ 但是加注释后"//"自动放在行首,没缩进,就像这样

  6. 转Struts 权限控制

    权限最核心的是业务逻辑,具体用什么技术来实现就简单得多. 通常:用户与角色建立多对多关系,角色与业务模块构成多对多关系,权限管理在后者关系中. 对权限的拦截,如果系统请求量大,可以用Struts2拦截 ...

  7. onlineDDL测试

    onlineDDL语法: alter table ALTER [COLUMN] col_name {SET DEFAULT literal | DROP DEFAULT} ADD [COLUMN] c ...

  8. [HIHO1318]非法二进制(动态规划)

    题目链接:http://hihocoder.com/problemset/problem/1318 题意:是个dp题.考虑二进制数为i位的时候,无非有两种情况:新添加的一位为0或者1. 为0的时候,那 ...

  9. [POJ2828]Buy Tickets(线段树,单点更新,二分,逆序)

    题目链接:http://poj.org/problem?id=2828 由于最后一个人的位置一定是不会变的,所以我们倒着做,先插入最后一个人. 我们每次处理的时候,由于已经知道了这个人的位置k,这个位 ...

  10. 【转】android UI设计的一些心得与问题解决(无效果图)

    1.把Button或者ImageButton的背景设为透明或者半透明: 半透明<Buttonandroid:background="#e0000000" ... /> ...