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// Fossil Fuels
// 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 2100000
int N,S,M;
int sz;
LL treeD[LIM],treeS[LIM];
int lazyM[LIM];
void Prop(int i)
{
int m=lazyM[i];
if (m)
{
lazyM[i]=0;
treeS[i]=treeD[i]+m;
if (i<sz)
{
i<<=1;
lazyM[i]=lazyM[i+1]=m;
}
}
}
void UpdateD(int i,int r1,int r2,int a,LL d)
{
Prop(i);
if ((r1==a) && (r2==a))
{
treeD[i]=d;
return;
}
int m=(r1+r2)>>1;
int c=i<<1;
if (a<=m)
UpdateD(c,r1,m,a,d);
else
UpdateD(c+1,m+1,r2,a,d);
Prop(c),Prop(c+1);
treeD[i]=min(treeD[c],treeD[c+1]);
treeS[i]=min(treeS[c],treeS[c+1]);
}
void UpdateM(int i,int r1,int r2,int a,int b,int mm)
{
Prop(i);
if ((a<=r1) && (r2<=b))
{
lazyM[i]=mm;
Prop(i);
return;
}
int m=(r1+r2)>>1;
int c=i<<1;
if (a<=m)
UpdateM(c,r1,m,a,b,mm);
if (b>m)
UpdateM(c+1,m+1,r2,a,b,mm);
Prop(c),Prop(c+1);
treeS[i]=min(treeS[c],treeS[c+1]);
}
LL QueryS(int i,int r1,int r2,int a,int b)
{
Prop(i);
if ((a<=r1) && (r2<=b))
return(treeS[i]);
LL ret=(LL)INF*INF;
int m=(r1+r2)>>1;
int c=i<<1;
if (a<=m)
Min(ret,QueryS(c,r1,m,a,b));
if (b>m)
Min(ret,QueryS(c+1,m+1,r2,a,b));
return(ret);
}
int main()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int K,L,A,X,Y,Z;
int i,j,k,m;
LL d;
static PR P[LIM];
static LL dyn[LIM];
stack<int> MD;
// testcase loop
Read(T);
Fox1(t,T)
{
// input
Read(N),Read(S),Read(M),Read(K);
k=0;
Fox(i,K*2)
{
if (i==K)
k=0;
Read(L),Read(A),Read(X),Read(Y),Read(Z);
Fox(j,L)
{
if (i<K)
P[k++].x=A;
else
P[k++].y=A;
A=((LL)X*A+Y)%Z+1;
}
}
// sort deposits by position
sort(P,P+N);
// init segment tree
for(sz=1;sz<N+1;sz<<=1);
Fill(treeD,60);
Fill(treeS,60);
Fill(lazyM,0);
// DP
while (!MD.empty())
MD.pop();
MD.push(-1);
dyn[0]=0;
UpdateD(1,0,sz-1,0,0);
j=0;
Fox(i,N)
{
// update tree back to index of previous larger depth
while ((MD.size()>1) && (P[MD.top()].y<=P[i].y))
MD.pop();
UpdateM(1,0,sz-1,MD.top()+1,i,P[i].y);
MD.push(i);
// advance left marker while it's too far away
while (P[j].x<P[i].x-2*M)
j++;
// query optimal DP value
d=S+QueryS(1,0,sz-1,j,i);
dyn[i+1]=d;
// insert DP value into tree
UpdateD(1,0,sz-1,i+1,d);
}
// output
printf("Case #%d: %lld\n",t,dyn[N]);
}
return(0);
} |