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Maze
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1419 Accepted Submission(s): 511
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
有一颗树n个结点n-1条边，根结点为1
对于在点i下一步有3种情况：
1：被杀死回到点1 — 概率为ki
2：找到出口退出—-慨率为ei
3：沿着边进入下一个点
求从点1開始到退出的平均须要走的边数
/*分析：
对于点i：
1,点i是叶子结点，则：
E(i)=ki*E(1)+ei*0+(1-ki-ei)*(E(father)+1)
=>E(i)=ki*E(1)+(1-ki-ei)*E(father)+(1-ki-ei)
2,点i非叶子结点，则：
E(i)=ki*E(1)+ei*0+(1-ki-ei)/m *(E(father)+1)+(1-ki-ei)/m*SUM(E(child)+1)
=>E(i)=ki*E(1)+(1-ki-ei)/m *E(father)+(1-ki-ei)/m*SUM(E(child))+(1-ki-ei);//作为1式
从公式可知求E(i)须要求到E(father),E(child)
但这是非常难求到的，由于即使是叶子结点也须要知道E(1),可是E(1)是未知的须要求的
如果:E(i)=Ai*E(1)+Bi*E(father)+Ci;//作为2式
所以:E(child)=Aj*E(1)+Bj*E(i)+Cj;
=>SUM(E(child))=SUm(Aj*E(1)+Bj*E(i)+Cj);
带入1式
=>E(i)=ki*E(1)+(1-ki-ei)/m *E(father)+(1-ki-ei)/m*SUm(Aj*E(1)+Bj*E(i)+Cj)+(1-ki-ei);
=>(1-(1-ki-ei)/m*SUM(Bj))*E(i)=(ki+(1-ki-ei)/m*SUM(Aj))*E(1)+(1-ki-ei)/m *E(father)+(1-ki-ei+(1-ki-ei)/m*SUM(cj));
与上述2式对照得到:
Ai=(ki+(1-ki-ei)/m*SUM(Aj)) / (1-(1-ki-ei)/m*SUM(Bj))
Bi=(1-ki-ei)/m / (1-(1-ki-ei)/m*SUM(Bj))
Ci=(1-ki-ei+(1-ki-ei)/m*SUM(cj)) / (1-(1-ki-ei)/m*SUM(Bj))
所以Ai,Bi,Ci仅仅与i的孩子Aj,Bj,Cj和本身ki,ei有关
于是能够从叶子開始逆推得到A1,B1,C1
在叶子节点：
Ai=ki;
Bi=(1-ki-ei);
Ci=(1-ki-ei);
而E(1)=A1*E(1)+B1*0+C1;
=>E(1)=C1/(1-A1);
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=10000+10;
const double eps=1e-9;
int n,size;
int head[MAX];
double A,B,C,k[MAX],e[MAX];
struct Edge{
int v,next;
Edge(){}
Edge(int V,int NEXT):v(V),next(NEXT){}
}edge[MAX*2];
void Init(){
memset(head,-1,sizeof head);
size=0;
}
void InsertEdge(int u,int v){
edge[size]=Edge(v,head[u]);
head[u]=size++;
}
void dfs(int u,int father){
double a=0,b=0,c=0,p;
int m=0;
for(int i=head[u]; i != -1;i=edge[i].next){
int v=edge[i].v;
if(v == father)continue;
dfs(v,u);
a+=A;
b+=B;
c+=C;
++m;
}
if(father != -1)++m;
p=(1-k[u]-e[u])/m;
A=(k[u]+p*a)/(1-p*b);
B=p/(1-p*b);
C=(1-k[u]-e[u]+p*c)/(1-p*b);
}
int main(){
int t,u,v,num=0;
scanf("%d",&t);
while(t--){
scanf( "%d",&n);
Init();
for(int i=1;i<n;++i){
scanf("%d%d",&u,&v);
InsertEdge(u,v);
InsertEdge(v,u);
}
for(int i=1;i<=n;++i){
scanf("%lf%lf",&k[i],&e[i]);
k[i]/=100;
e[i]/=100;
}
dfs(1,-1);
if(fabs(A-1)<eps)printf("Case %d: impossible\n",++num);
else printf("Case %d: %.6f\n",++num,C/(1-A));
}
return 0;
}
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