bzoj1075

题意:

给你一个地图，问从x1,y1->x2,y2，要走的路最短，问

耗油和速度

题解：

首先把他们转到左下角->右上角

然后只能往上或往下

考虑到可能有小数

所以都乘上他们的公倍数

然后就是dp

代码：

#include<bits/stdc++.h>
using namespace std;
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 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 (int i=1;i<=10;i++)w[i]=1.0*L/(80.0-0.75*i*i);
      for (int i=1;i<=n;i++)scanf("%d",&a[i]),a[i]/=5;
      for (int i=1;i<=n;i++)scanf("%d",&b[i]),b[i]/=5;
      for (int i=1;i<=n;i++)x=max(max(a[i],b[i]),x);
      for (int 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 (int i=1,j=n;i<j;i++,j--)swap(a[i],a[j]);
        ys=n-ys+1,yt=n-yt+1;
       }
      for (int j=ys;j<=yt;j++)
     for (int k=0;k<=lim;k++)f[0][j][k]=inf;
      f[0][ys][0]=0;
      for (int i=xs;i<=xt;i++,p^=1)
     {
        for (int j=ys;j<=yt;j++)
         for (int k=0;k<=lim;k++)f[p^1][j][k]=inf;
        for (int j=ys;j<=yt;j++)
         for (int k=0;k<=lim;k++)
          if (f[p][j][k]<inf)
           {
              if (j<yt)
             for (int 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 (int 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 (int 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)
     {
         puts("No");
         return 0;
     }
      printf("%d %.2f\n%d %.2f",cal(ans1),f[p^1][yt][ans1],
    cal(ans2),f[p^1][yt][ans2]);
      return 0;
}