标签：ever else elf head mes restart error include accept
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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=1e4+5;
const double eps=1e-9;
int T,n;
int deg[N];
double A[N],B[N],C[N];
double k[N],e[N];
int head[N],num_edge;
struct Edge
{
int v,nxt;
}edge[N<<1];
inline void add_edge(int u,int v)
{
edge[++num_edge].v=v;
edge[num_edge].nxt=head[u];
head[u]=num_edge;
}
bool dfs(int u,int fa)
{
double m=deg[u];
A[u]=k[u];
B[u]=(1-k[u]-e[u])/m;
C[u]=1-k[u]-e[u];
double tmp=1,ratio=B[u];
for(int i=head[u],v;i;i=edge[i].nxt)
{
v=edge[i].v;
if(v==fa)
continue;
if(!dfs(v,u))
return false;
tmp-=ratio*B[v];
A[u]+=ratio*A[v];
C[u]+=ratio*C[v];
}
if(fabs(tmp)<=eps)
return false;
A[u]/=tmp;
B[u]/=tmp;
C[u]/=tmp;
return true;
}
int taskid;
int main()
{
scanf("%d",&T);
while(T--)
{
memset(deg,0,sizeof(deg));
memset(head,0,sizeof(head));
num_edge=0;
scanf("%d",&n);
for(int i=1,a,b;i<n;++i)
{
scanf("%d%d",&a,&b);
++deg[a],++deg[b];
add_edge(a,b);
add_edge(b,a);
}
for(int i=1;i<=n;++i)
{
scanf("%lf%lf",k+i,e+i);
k[i]/=100,e[i]/=100;
}
cout<<"Case "<<++taskid<<": ";
if(dfs(1,1)&&fabs(1-A[1])>eps)
cout<<C[1]/(1-A[1])<<‘\n‘;
else
puts("impossible");
}
return 0;
}
Hdu 4035 Maze(概率DP)
标签：ever else elf head mes restart error include accept
原文地址：https://www.cnblogs.com/lovewhy/p/9788362.html
