// RNG // Official Solution by Jacob Plachta #define DEBUG 0 #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; i=0; i--) #define FoxR1(i,n) for (i=n; i>0; i--) #define FoxRI(i,a,b) for (i=b; i>=a; i--) #define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++) #define Min(a,b) a=min(a,b) #define Max(a,b) a=max(a,b) #define Sz(s) int((s).size()) #define All(s) (s).begin(),(s).end() #define Fill(s,v) memset(s,v,sizeof(s)) #define pb push_back #define mp make_pair #define x first #define y second template T Abs(T x) { return(x<0 ? -x : x); } template T Sqr(T x) { return(x*x); } string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); } const int INF = (int)1e9; const LD EPS = 1e-9; const LD PI = acos(-1.0); //#if DEBUG #define GETCHAR getchar /*#else #define GETCHAR getchar_unlocked #endif*/ bool Read(int &x) { char c,r=0,n=0; x=0; for(;;) { c=GETCHAR(); if ((c<0) && (!r)) return(0); if ((c=='-') && (!r)) n=1; else if ((c>='0') && (c<='9')) x=x*10+c-'0',r=1; else if (r) break; } if (n) x=-x; return(1); } int main() { if (DEBUG) freopen("in.txt","r",stdin); int TT,tt; int N,M,K,T,R; LD P; int i,j,k,b; LD d,v; int S[20]; static int up[100000],down[100000],dist[100000],dir[20][20]; static vector con[100000]; static LD exRN[100001],exNR[100001],exP[100001],exT[100001]; static LD dyn[1<<20][21]; queue Q; Read(TT); Fox1(tt,TT) { //input Read(N),Read(M),Read(K),Read(T),Read(R); Fox(i,N) con[i].clear(); cin >> P; Fox(i,K) Read(S[i]),S[i]--; while (M--) { Read(i),Read(j); con[i-1].pb(j-1); } Fox(i,K) dyn[(1< node exRN[0]=0; Fox1(i,N) { v=(P*T + (1-P)*(R+exRN[i-1])) / P; exRN[i]=exRN[i-1]+v; } //expected times for node -> root exNR[0]=R; Fox1(i,N) exNR[i]=P*(T + exNR[i-1]) + (1-P)*R; //conditional expected times for node -> node exP[0]=1; exT[0]=0; Fox1(i,N) { exP[i]=exP[i-1]*P; exT[i]=exT[i-1] + exP[i-1]*(1-P)*((i-1)*T + R); } //bitmask DP FoxR(b,(1<v)) d=v; } dyn[b][K]=d; Fox(i,K) if (b&(1<=0) { k=dir[i][j]; v=exP[k]*(k*T + dyn[b][j]) + exT[k] + (1-exP[k])*dyn[b][K]; Min(d,v); } dyn[b][i]=d; } } //output printf("Case #%d: ",tt); cout << fixed << setprecision(9) << dyn[0][K] << endl; } return(0); }