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3.5 kB
| // Graph Gift | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| 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); | |
| 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 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); | |
| } |