2019/round3/renovations.cpp

// Renovations
// Solution by Jacob Plachta

#define DEBUG 0

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; 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<typename T> T Abs(T x) { return(x<0 ? -x : x); }
template<typename T> 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-12;
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);
}

#define LIM 6001
#define MOD 1000000007

int N,K;
int expMem[LIM];

int Exp(int a,int b)

{
		if ((a<N) && (b==K) && (expMem[a]))
			return(expMem[a]);
	LL p=a,v=1;
		while (b)
		{
				if (b&1)
					v=v*p%MOD;
			p=p*p%MOD;
			b>>=1;
		}
		if ((a<N) && (b==K))
			expMem[a]=v;
	return(v);
}

void Add(int &a,int b)

{
	a=(a+b)%MOD;
}

int Prod(int a,int b)

{
	return((LL)a*b%MOD);
}

void Div(int &a,int b)

{
	a=Prod(a,Exp(b,MOD-2));
}

// # of sequences with f forbidden nodes, r required nodes
int Count(int f,int r)

{
	int v=N-f-1;
	int c=Exp(v,K);
		if (!r)
			return(c);
		if (r==1)
			return((c-Exp(v-1,K)+MOD)%MOD);
	return((c-(LL)2*Exp(v-1,K)%MOD+Exp(v-2,K)+MOD)%MOD);
}

int main()

{
		if (DEBUG)
			freopen("in.txt","r",stdin);
	// vars
	int T,t;
	int L,A,B;
	int i,dA,dB,dL;
	int P[LIM]={-1};
	// testcase loop
	Read(T);
		Fox1(t,T)
		{
			// init
			Fill(expMem,0);
			// input
			Read(N),Read(K),Read(A),Read(B),A--,B--;
				Fox1(i,N-1)
					Read(P[i]),P[i]--;
			// compute depths of A, B, and their LCA
			queue<int> qA,qB;
				for (dA=0; A>=0; A=P[A],dA++)
					qA.push(A);
				for (dB=0; B>=0; B=P[B],dB++)
					qB.push(B);
			dL=max(dA,dB);
				for(;;)
				{
						if (qA.front()==qB.front())
							break;
					dL--;
						if (Sz(qA)>dL)
							qA.pop();
						if (Sz(qB)>dL)
							qB.pop();
				}
			dL--,dA--,dB--;
			// consider all possible final states
			int ans=0;
				if (!dL) // original LCA is the root?
				{
					Fox(A,dA+1)
						Fox(B,dB+1)
						{
								if ((dA) && (!A) || (dB) && (!B))
									continue;
							Add(ans,Prod(A+B,Count(A+B-(A?1:0)-(B?1:0),(A<dA?1:0)+(B<dB?1:0))));
						}
				}
				else
				{
					// consider LCA being retained
						Fox1(L,dL)
							Add(ans,Prod(dA+dB-2*dL,Count(dA+dB-2*dL+L-1,(L<dL ? 1 : 0))));
					// consider LCA being broken
						Fox1(A,dA)
							Fox1(B,dB)
							{
									if ((A>dA-dL) && (B>dB-dL))
										continue;
								Add(ans,Prod(A+B,Count(A+B-2,(A<dA?1:0)+(B<dB?1:0))));
							}
				}
			// divide by total number of sequences
			Div(ans,Exp(N-1,K));
			// output
			printf("Case #%d: %d\n",t,ans);
		}
	return(0);
}