bzoj4720

这道是NOIP的题目。

先用SPFA找出所有最短路

用f[i][j]表示到第i个时间段时决定提出申请换教室，而加上这次申请总共用了j次申请的最小期望

而g则表示这次不申请。

直接按照期望的情况（申请成功与否去转移就好了）

#include<cstdio>
#include<cstdlib>
#include<cstring>
int n,m,v,E;
int c[30005];
int d[30005];
double p[30005];
struct mod{int x,y,z,next;};
mod q[400005];
int first[30005];
int dis[505][505];
int t[30005];
bool vis[3005];
double g[2005][2005];
double f[2005][2005];
int len=0;
void ins(int x,int y,int z)
{
    len++;
    q[len].x=x;
    q[len].y=y;
    q[len].z=z;
    q[len].next=first[x];
    first[x]=len;
}
double mymin(double u1,double u2)
{
 if (u1<u2)return u1;
 return u2; 
}
void SPFA(int o)
{
 int st=1,ed=2; 
 memset(vis,false,sizeof(v));
 vis[o]=true;
 t[1]=o;
 while(st!=ed)
 {
  int x=t[st];
  for (int i=first[x];i!=0;i=q[i].next)
  {
   int y=q[i].y;
   if (dis[o][x]+q[i].z<dis[o][y])
   {
    dis[o][y]=dis[o][x]+q[i].z;
    if (vis[y]==false)
    {
     vis[y]=true;
     t[ed]=y;
     ed++;
     if (ed>v)ed=1;  
    }
   }
  }
  vis[x]=false;
  st++;
  if (st>v)st=1;
 }
}
int main()
{
    scanf("%d%d%d%d",&n,&m,&v,&E);
    for (int i=1;i<=n;i++)
    scanf("%d",&c[i]);
    for (int i=1;i<=n;i++)
    scanf("%d",&d[i]);
    for (int i=1;i<=n;i++)
    scanf("%lf",&p[i]);
    memset(first,0,sizeof(first));
    for (int i=1;i<=E;i++)
    {
     int x,y,z;
     scanf("%d%d%d",&x,&y,&z);
     ins(x,y,z);
     ins(y,x,z);
    }
    memset(dis,63,sizeof(dis));
    for (int i=1;i<=v;i++)
    {
     dis[i][i]=0;
     SPFA(i); 
    }
    /*
    for (int i=1;i<=v;i++)
      for (int j=1;j<=v;j++)
      {
       printf("i:%d j:%d d:%d\n",i,j,dis[i][j]);    
      }
     */
    for (int i=0;i<=n;i++)
      for (int j=0;j<=m;j++)
      {
      f[i][j]=999999999.0;
      g[i][j]=999999999.0;  
      }
    f[1][1]=0;g[1][0]=0;
    for (int i=2;i<=n;i++)
      for (int j=0;j<=mymin(m,i);j++)
      {
       g[i][j]=mymin(g[i][j],g[i-1][j]+dis[c[i-1]][c[i]]);
       if (j>=1)
       {
       g[i][j]=mymin(
                    g[i][j],
                    (f[i-1][j]+dis[d[i-1]][c[i]])*p[i-1]+(f[i-1][j]+dis[c[i-1]][c[i]])*(1.0-p[i-1])
                    );
       f[i][j]=mymin(
                    f[i][j],
                    (g[i-1][j-1]+dis[c[i-1]][d[i]])*p[i]+(g[i-1][j-1]+dis[c[i-1]][c[i]])*(1.0-p[i])
                    );
       }
       if (j>1)
       {
       f[i][j]=mymin(
                    f[i][j],
                    (f[i-1][j-1]+dis[c[i-1]][c[i]])*(1.0-p[i-1])*(1.0-p[i])+
                    (f[i-1][j-1]+dis[c[i-1]][d[i]])*(1.0-p[i-1])*p[i]+
                    (f[i-1][j-1]+dis[d[i-1]][c[i]])*p[i-1]*(1.0-p[i])+
                    (f[i-1][j-1]+dis[d[i-1]][d[i]])*p[i-1]*p[i]
                    );
       }
      }
      double ans=999999999.0;
      for (int i=0;i<=m;i++)
      ans=mymin(ans,mymin(f[n][i],g[n][i]));
      printf("%.2lf\n",ans);
}