HDU-4035 Maze 概率DP
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4035
很不错的概率DP题目,因为这题是无向图,所以要对叶节点和非叶节点考虑,然后列出方程后,因为数据很大,高斯消元如果不特定优化会超时,可以转化方程,然后求解系数。
解法:<摘自KB神>
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,开始走,在每个结点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;
从叶子结点开始,直到算出 A1,B1,C1;
E[1] = A1*E[1] + B1*0 + C1;
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...
1 //STATUS:C++_AC_281MS_1440KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=10010; 37 const int INF=0x3f3f3f3f; 38 const int MOD=10007,STA=8000010; 39 const LL LNF=1LL<<55; 40 const double EPS=1e-9; 41 const double OO=1e30; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 vector<int> q[N]; 59 double k[N],e[N],A[N],B[N],C[N]; 60 int T,n; 61 62 void dfs(int u,int fa) 63 { 64 int i,j,v,m=q[u].size(); 65 double P=(1-k[u]-e[u])/(m); 66 double At,Bt,Ct; 67 At=Bt=Ct=0; 68 for(i=0;i<m;i++){ 69 v=q[u][i]; 70 if(v==fa)continue; 71 dfs(v,u); 72 At+=A[v]; 73 Bt+=B[v]; 74 Ct+=C[v]; 75 } 76 A[u]=(P*At+k[u])/(1-P*Bt); 77 B[u]=P/(1-P*Bt); 78 C[u]=(P*Ct+1-k[u]-e[u])/(1-P*Bt); 79 } 80 81 int main(){ 82 // freopen("in.txt","r",stdin); 83 int ca=1,i,j,a,b; 84 scanf("%d",&T); 85 while(T--) 86 { 87 scanf("%d",&n); 88 for(i=1;i<=n;i++)q[i].clear(); 89 for(i=1;i<n;i++){ 90 scanf("%d%d",&a,&b); 91 q[a].push_back(b); 92 q[b].push_back(a); 93 } 94 for(i=1;i<=n;i++){ 95 scanf("%lf%lf",&k[i],&e[i]); 96 k[i]/=100,e[i]/=100; 97 } 98 99 dfs(1,0); 100 101 printf("Case %d: ",ca++); 102 if(sign(A[1]-1))printf("%.6lf\n",C[1]/(1-A[1])); 103 else printf("impossible\n"); 104 } 105 return 0; 106 }