HDU-4035 Maze （概率DP&&树形DP）
Maze
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
题目大意：给定一个树状迷宫，初始在1号结点，在i号结点被杀死的概率为：k[i]（百分比表示），走出迷宫的概率为：e[i]（百分比表示），若未走出迷宫且未被杀死，则等可能的走向与其相连的一个结点，求走出这个迷宫的概率？
理解题以后觉得没法做。。。只理解了有向图，因为本题是无向边，所以父子结点的概率互相影响，立刻就忘了上一题也存在这种情况。。。
看了题解后才明白，应该按照期望设状态，而且需要在树上进行状态转移，最开始的状态转移方程最后化简即可消除影响（貌似就是树形DP+概率DP）
设dp[i]表示从i号结点走出迷宫时走过边数的概率，par[i]表示i号结点的父亲结点，son表示i号结点的儿子结点：
①若i结点为叶子结点，则 dp[i]=k[i]*dp[1]+e[i]*0+(1-k[i]-e[i])*(dp[par[i]]+1);
②若i结点不为叶子结点，则dp[i]=k[i]*dp[1]+e[i]*0+(1-k[i]-e[i])*((1/m)*(dp[par[i]]+1)+∑((1/m)*(dp[son[i]]+1)));【m表示与i相连的结点数】
由于每个结点都和1号结点以及其父亲结点有线性关系，所以设参数a[i],b[i],c[i]，令dp[i]=a[i]*dp[1]+b[i]*dp[par[i]]+c[i];
将上式带入状态转移方程两式的可得：
①若i结点为叶子结点，则对比参数可得：
a[i]=k[i];
b[i]=1-k[i]-e[i];
c[i]=1-k[i]-e[i];
②若i结点不为叶子结点，设j为i结点的叶子结点：
则状态转移方程为：dp[i]=k[i]*dp[1]+e[i]*0+(1-k[i]-e[i])*((1/m)*(dp[par[i]]+1)+∑((1/m)*(dp[j]+1)));【m表示与i相连的结点数】
代入后得：dp[i]=k[i]*dp[1]+(1-k[i]-e[i])*((1/m)*(dp[par[i]]+1)+∑((1/m)*(a[j]*dp[1]+b[j]*dp[par[j]]+c[j]+1)));
化简合并同类项得：(1-(1-k[i]-e[i])/m*∑b[j])*dp[i]=(k[i]+(1-k[i]-e[i])/m*∑a[j])*dp[1]+(1-k[i]-e[i])/m)*dp[par[i]]+(1-k[i]-e[i])/m)*∑c[j]+1-k[i]-e[i];
对比参数可得：
a[i]= (k[i]+(1-k[i]-e[i])/m*∑a[j]) / (1-(1-k[i]-e[i])/m*∑b[j]) ;
b[i]= (1-k[i]-e[i])/m) / (1-(1-k[i]-e[i])/m*∑b[j]) ;
c[i]= ((1-k[i]-e[i])/m)*∑c[j]+(1-k[i]-e[i])) / (1-(1-k[i]-e[i])/m*∑b[j]) ;
而dp[1]=a[1]*dp[1]+b[1]*0+c[1] 可以推出：dp[1]=c[1]/(1-a[1]);
所以从叶子结点一次算出a[i],b[i],c[i]，再求出dp[1];
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN=10005;
const double EPS=1e-9;
int n,k1,k2,k3,x,y,z;
double a[MAXN],b[MAXN],c[MAXN],k[MAXN],e[MAXN],sum;
vector<int> g[MAXN];
bool dfs(int u,int par) {
int m=g[u].size();
double tt=0,tmp;
a[u]=k[u];
tmp=b[u]=(1-k[u]-e[u])/m;//若u为叶子结点，则m==1，所以可以将所有的都初始化为该值
c[u]=1-k[u]-e[u];
int v;
for(int i=0;i<g[u].size();++i) {
v=g[u][i];
if(v!=par) {
if(!dfs(v,u)) {
return false;
}
a[u]+=tmp*a[v];
c[u]+=tmp*c[v];
tt+=tmp*b[v];
}
}
if(abs(1-tt)<EPS) {//若1-tt趋近于0，则无解
return false;
}
a[u]/=1-tt;
b[u]/=1-tt;
c[u]/=1-tt;
return true;
}
int main() {
int T,kase=0,s,f;
scanf("%d",&T);
while(kase<T) {
scanf("%d",&n);
for(int i=1;i<=n;++i) {
g[i].clear();
}
for(int i=1;i<n;++i) {
scanf("%d%d",&s,&f);
g[s].push_back(f);
g[f].push_back(s);
}
for(int i=1;i<=n;++i) {
scanf("%lf%lf",k+i,e+i);
k[i]/=100;
e[i]/=100;
}
printf("Case %d: ",++kase);
if(dfs(1,0)&&abs(1-a[1]>EPS)) {//若1-a[1]趋近于0，则无解
printf("%.6lf\n",c[1]/(1-a[1]));
}
else {
printf("impossible\n");
}
}
return 0;
}