Maze
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1677 Accepted Submission(s): 638
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.
from the maze, output “impossible”.
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
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=10010;
const double eps=1e-10;//开始-6wa了。精度要高点才行。
typedef long long ll;
struct node
{
int v;
node *next;
} ed[maxn<<1],*head[maxn];
int cnt,eds[maxn];
double A[maxn],B[maxn],C[maxn],ki[maxn],ei[maxn];
void adde(int u,int v)
{
ed[cnt].v=v;
ed[cnt].next=head[u];
head[u]=&ed[cnt++];
}
void dfs(int fa,int u)
{
double sa,sb,sc,mi;
sa=sb=sc=0;
for(node *p=head[u];p!=NULL;p=p->next)
{
if(p->v==fa)
continue;
dfs(u,p->v);
sa+=A[p->v];
sb+=B[p->v];
sc+=C[p->v];
}
mi=(1-ki[u]-ei[u])/eds[u];
A[u]=(ki[u]+mi*sa)/(1-mi*sb);
B[u]=mi/(1-mi*sb);
C[u]=(1+mi*sc-ki[u]-ei[u])/(1-mi*sb);
}
int main()
{
int t,cas=1,n,i,u,v;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
cnt=0;
memset(head,0,sizeof head);
memset(eds,0,sizeof eds);
for(i=1;i<n;i++)
{
scanf("%d%d",&u,&v);
eds[u]++,eds[v]++;
adde(u,v);
adde(v,u);
}
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&ki[i],&ei[i]);
ki[i]/=100,ei[i]/=100;
}
dfs(-1,1);
printf("Case %d: ",cas++);
if(fabs(1-A[1])<eps)
printf("impossible\n");
else
printf("%f\n",C[1]/(1-A[1]));
}
return 0;
}