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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; }
BZOJ1075 : [SCOI2007]最优驾车drive
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原文地址:http://www.cnblogs.com/clrs97/p/5029473.html