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// Graph Gift
// 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 30003
int NA,NB;
LL A[LIM],B[LIM];
LL SB[LIM];
LL mem3[LIM];
// cost for cross-pairing B values such that ny = n-y
LL Comp3(int ny)
{
if (mem3[ny])
return(mem3[ny]);
int i;
LL ans=0;
Fox(i,(ny-1)/2)
ans+=B[NB-i-2]*B[NB-ny+i];
return(mem3[ny]=ans);
}
// min. cost for connecting last B values
LL Comp2(int n)
{
if (!n)
return(0);
if ((n==1) && (NB>=2))
return(B[NB-1]*B[NB-2]);
int y;
LL cur,ans=(LL)6e18;
// pair first y values with the last value
Fox1(y,n-1)
{
if ((n-y-1)%2)
continue;
cur=B[NB-1]*(SB[NB-n+y]-SB[NB-n]);
// cross-pair the remaining n-y-1 values
cur+=Comp3(n-y);
Min(ans,cur);
}
return(ans);
}
// min. cost with last x A values paired with different B values
LL Comp1(int x)
{
int i;
LL ans=0;
Fox(i,NA-x)
ans-=B[0]*A[i];
Fox(i,x)
ans-=B[i+1]*A[NA-x+i];
ans+=Comp2(NB-x-1);
return(ans);
}
int main()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int N;
int i,x;
LL ans,v;
// testcase loop
Read(T);
Fox1(t,T)
{
// input
Read(N);
NA=NB=0;
while (N--)
{
Read(x);
if (x>0)
A[NA++]=x;
else
B[NB++]=-x;
}
// sort each list in non-increasing order
sort(A,A+NA);
sort(B,B+NB);
reverse(A,A+NA);
reverse(B,B+NB);
// precompute prefix sums
Fox(i,NB)
SB[i+1]=SB[i]+B[i];
// special case: all positive/negative?
Fill(mem3,0);
if ((!NA) || (!NB))
{
if (!NB)
{
NB=NA;
memcpy(B,A,sizeof(A));
Fox(i,NB)
SB[i+1]=SB[i]+B[i];
}
ans=Comp2(NB);
}
else
{
// try all parameters
ans=(LL)6e18;
Fox(x,min(NA,NB))
Min(ans,Comp1(x));
}
// output
printf("Case #%d: %lld\n",t,ans);
}
return(0);
} |