// Ethan Sums Shortest Distances // Solution by Jacob Plachta #define DEBUG 0 #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define LL long long #define LD long double #define PR pair #define Fox(i,n) for (i=0; 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 T Abs(T x) { return(x<0 ? -x : x); } template 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 51 int R[2][LIM],sum[2][LIM]; LL dyn[LIM][3][LIM]; // dyn[i][r][p] = min. cost such that: // - you're ending at a vertical edge in column i (its cost is exluded) // - you previously had a partial horizontal section in row r (r=2 indicates both rows) // - the partial horizontal section started in column p int main() { if (DEBUG) freopen("in.txt","r",stdin); // vars int T,t; int N; LL S; int i,i2,j,k,r,r2,p,p2,s; LL ans,cur,cur2; // testcase loop Read(T); Fox1(t,T) { // input, and compute each row's prefix sums Read(N); Fox(i,2) Fox(j,N) { Read(R[i][j]); sum[i][j+1]=sum[i][j]+R[i][j]; } S=sum[0][N]+sum[1][N]; // initial DP step (before first vertical edge) Fill(dyn,60); Fox(i,N) { cur=0; // compute horizontal section costs for both rows' prefixes Fox(j,2) { s=0; Fox(k,i) { s+=R[j][k]; cur+=s*(S-s); } } dyn[i][2][0]=cur; } // main DP Fox(i,N) FoxI(i2,i+1,N-1) Fox(r2,2) FoxI(p2,i+1,i2) { cur=0; // compute full horizontal section cost s=sum[r2][p2]+sum[1-r2][i+1]; FoxI(j,i,i2-1) { cur+=s*(S-s); s+=R[1-r2][j+1]; } // compute left partial horizontal section cost s=0; FoxRI(j,i+1,p2-1) { s+=R[r2][j]; cur+=s*(S-s); } // compute right partial horizontal section cost s=0; FoxI(j,p2,i2-1) { s+=R[r2][j]; cur+=s*(S-s); } // consider all previous states Fox(r,3) Fox(p,i+1) { cur2=dyn[i][r][p]+cur; // compute vertical edge cost if (r==2) s=sum[r2][p2]; else if (r==r2) s=sum[r2][p2]-sum[r2][p]; else s=sum[r2][p2]+sum[1-r2][p]; cur2+=s*(S-s); Min(dyn[i2][r2][p2],cur2); } } // final DP step (after last vertical edge) ans=(LL)INF*INF; Fox(i,N) Fox(r,3) Fox(p,i+1) { cur=dyn[i][r][p]; // compute horizontal section costs for both rows' suffixes Fox(j,2) { s=0; FoxRI(k,i+1,N-1) { s+=R[j][k]; cur+=s*(S-s); } } // compute vertical edge cost if (r==2) s=sum[0][N]; else s=sum[r][N]-sum[r][p]; cur+=s*(S-s); Min(ans,cur); } // output printf("Case #%d: %lld\n",t,ans); } return(0); }