CodeForces596D Wilbur and Trees

Posted 2020-11-07 LJZ_C

题目链接

一个人要去砍一排n棵树，他会等概率选取当前最左边或最右边的树，树有p的概率向左倒，1-p的概率向右倒，给出每棵树在数轴上的位置与树的高度，求树覆盖长度的期望

dp[l][r][lk][rk]表示区间[l,r]与l-1棵树倒的方向，r+1棵树倒的方向
更新方法比较好想，细节比较繁琐

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2005;
int h;
double p;
int a[maxn];
double dp[maxn][maxn][2][2];
double DW(int l,int r,int sl,int sr) 
{
    if (dp[l][r][sl][sr]) return dp[l][r][sl][sr];
    if (l==r)
    {
        if (sl&&a[l-1]+h>a[l])
        {
            if (sr) return dp[l][r][sl][sr]=min(h,a[l+1]-a[l]);
            else return dp[l][r][sl][sr]=min(h,max(0,a[r+1]-a[r]-h));
        }
        else if (!sr&&a[r+1]-h<a[r])
        {
            if (sl) return dp[l][r][sl][sr]=min(h,max(a[l]-a[l-1]-h,0));
            else  return dp[l][r][sl][sr]=min(h,a[l]-a[l-1]);
        }
        double ans=0;
        if (sl) ans+=p*min(h,max(a[l]-a[l-1]-h,0));
        else ans+=p*min(h,a[l]-a[l-1]);
        if (sr) ans+=(1-p)*min(h,a[r+1]-a[r]);
        else ans+=(1-p)*min(h,max(0,a[r+1]-a[r]-h));
        return dp[l][r][sl][sr]=ans;
    }
    if (sl&&a[l-1]+h>a[l])      return 
    dp[l][r][sl][sr]=DW(l+1,r,1,sr)+min(h,a[l+1]-a[l]);
    else if(!sr&&a[r+1]-h<a[r]) return
    dp[l][r][sl][sr]=DW(l,r-1,sl,0)+min(h,a[r]-a[r-1]);
    double ans=0;
    ans+=0.5*(1-p)*(DW(l+1,r,1,sr)+min(h,a[l+1]-a[l]));
    ans+=0.5*p    *(DW(l,r-1,sl,0)+min(h,a[r]-a[r-1]));
    if (sl) ans+=0.5*p*(DW(l+1,r,0,sr)+min(h,max(0,a[l]-a[l-1]-h)));
    else    ans+=0.5*p*(DW(l+1,r,0,sr)+min(h,a[l]-a[l-1]));
    if (sr) ans+=0.5*(1-p)*(DW(l,r-1,sl,1)+min(h,a[r+1]-a[r]));
    else    ans+=0.5*(1-p)*(DW(l,r-1,sl,1)+min(h,max(0,a[r+1]-a[r]-h)));
    return dp[l][r][sl][sr]=ans;
}
int main()
{
    int n;
    while (~scanf("%d%d%lf",&n,&h,&p))
    {
        for (int i=1;i<=n;i++) scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        sort(a+1,a+n+1);
        a[0]=a[1]-h;
        a[n+1]=a[n]+h;
        printf("%.12lf\n",DW(1,n,0,1));
    }
    return 0;
}