Maze
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 1713 Accepted Submission(s): 659
Special Judge
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?
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.
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
Case 1: 2.000000 Case 2: impossible Case 3: 2.895522
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，開始走，在每一个结点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);
设对每一个结点：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;
从叶子结点開始，直到算出 A1,B1,C1;
E[1] = A1*E[1] + B1*0 + C1; //由于根节点无父节点
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...
</pre><pre name="code" class="cpp">#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
#define N 10005
#define LL __int64
vector<int>g[N];
const double eps=1e-10;
double a[N],b[N],c[N],e[N],k[N];
bool dfs(int t,int pre)
{
int i,m=g[t].size();
a[t]=k[t];
b[t]=(1-k[t]-e[t])/m;
c[t]=1-k[t]-e[t];
double tmp=0;
for(i=0;i<m;i++)
{
int v=g[t][i];
if(v==pre)
continue;
if(!dfs(v,t))
return false;
a[t]+=a[v]*(1-k[t]-e[t])/m;
c[t]+=c[v]*(1-k[t]-e[t])/m;
tmp+=b[v]*(1-k[t]-e[t])/m;
}
if(fabs(1-tmp)<eps)
return false;
a[t]/=(1-tmp);
b[t]/=(1-tmp);
c[t]/=(1-tmp);
return true;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int T,i,n,u,v,cnt=1;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=0;i<=n;i++)
g[i].clear();
for(i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&k[i],&e[i]);
k[i]/=100;
e[i]/=100;
}
printf("Case %d: ",cnt++);
if(dfs(1,-1)&&fabs(1-a[1])>eps)
{
double ans=c[1]/(1-a[1]);
printf("%.8f
",ans);
}
else
printf("impossible
");
}
return 0;
}