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// Connect the Dots
// 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 800001
int main()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int N,H,V;
int i,j,si,a,b,c,d;
static int G[2][LIM];
static PR P[LIM];
static int maxY[LIM];
set<PR> S;
set<PR>::iterator I;
// testcase loop
Read(T);
Fox1(t,T)
{
// input, and generate coordinates
Read(N),Read(H),Read(V);
Fox(i,2)
{
Read(G[i][0]),Read(G[i][1]),Read(a),Read(b),Read(c),Read(d);
FoxI(j,2,N-1)
G[i][j]=((LL)a*G[i][j-2] + (LL)b*G[i][j-1] + c)%d+1;
}
// impossible?
int ans=-1;
if (H+V<N)
goto Done;
// sort dots by X
Fox(i,N)
P[i]=mp(G[0][i],G[1][i]);
sort(P,P+N);
// precompute Y values' suffix maxes
maxY[N]=0;
FoxR(i,N)
maxY[i]=max(maxY[i+1],P[i].y);
// consider each possible max horiz. segment length
ans=2*INF;
S.clear();
S.insert(mp(0,-1));
I=S.begin(),si=0;
Fox(i,N+1)
{
// update set of dots so far
if (i)
{
S.insert(mp(P[i-1].y,i));
if (mp(P[i-1].y,i)<*I)
si++;
}
// last f dots must use vert. segments
int f=N-i;
if (f>V)
continue;
// g of the dots so far must also use vert. segments
int g=max(0,min(N-H,V)-f);
while (si>g)
I--,si--;
while (si<g)
I++,si++;
Min(ans,(i?P[i-1].x:0)+max(I->x,maxY[i]));
}
// output
Done:;
printf("Case #%d: %d\n",t,ans);
}
return(0);
} |