设$f[i][j][k]$为到达$(i,j)$，用时为$\frac{k}{5lcm}$小时的最低耗油量，然后DP即可。
#include<cstdio>
const int N=12,M=210005;
const double inf=1e15;
int n,L,lcm,lim,i,j,k,p,x,y,a[N],b[N],xs,ys,xt,yt,t1,t2,ans1=-1,ans2;
double f[2][N][M],w[N];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
void swap(int&a,int&b){int c=a;a=b;b=c;}
inline void up(double&a,double b){if(a>b)a=b;}
int cal(int x){
x*=12;
return x/lcm+(x%lcm>0);
}
int main(){
scanf("%d%d",&n,&L);
for(i=1;i<=10;i++)w[i]=1.0*L/(80.0-0.75*i*i);
for(i=1;i<=n;i++)scanf("%d",&a[i]),a[i]/=5;
for(i=1;i<=n;i++)scanf("%d",&b[i]),b[i]/=5;
for(i=1;i<=n;i++){
if(x<a[i])x=a[i];
if(x<b[i])x=b[i];
}
for(i=lcm=1;i<=x;i++)lcm=lcm*i/gcd(lcm,i);
scanf("%d%d%d%d%d%d",&xs,&ys,&xt,&yt,&t1,&t2);
lim=t2*lcm/12;
if(xs>xt)swap(xs,xt),swap(ys,yt);
if(ys>yt){
for(i=1,j=n;i<j;i++,j--)swap(a[i],a[j]);
ys=n-ys+1,yt=n-yt+1;
}
for(j=ys;j<=yt;j++)for(k=0;k<=lim;k++)f[0][j][k]=inf;
f[0][ys][0]=0;
for(i=xs;i<=xt;i++,p^=1){
for(j=ys;j<=yt;j++)for(k=0;k<=lim;k++)f[p^1][j][k]=inf;
for(j=ys;j<=yt;j++)for(k=0;k<=lim;k++)if(f[p][j][k]<inf){
if(j<yt)for(x=b[i];x;x--){
y=k+lcm/x*L;
if(y<=lim)up(f[p][j+1][y],f[p][j][k]+w[x]);
}
if(i<xt)for(x=a[j];x;x--){
y=k+lcm/x*L;
if(y<=lim)up(f[p^1][j][y],f[p][j][k]+w[x]);
}
}
}
for(k=0;k<=lim;k++)if(k*12>=t1*lcm&&f[p^1][yt][k]<inf){
if(ans1<0)ans1=k;
if(!ans2||f[p^1][yt][k]+1e-9<f[p^1][yt][ans2])ans2=k;
}
if(ans1<0)return puts("No"),0;
printf("%d %.2f\n%d %.2f",cal(ans1),f[p^1][yt][ans1],cal(ans2),f[p^1][yt][ans2]);
return 0;
}