CodeForces 629D Babaei and Birthday Cake（线段树维护DP）_a

本文链接：https://blog.youkuaiyun.com/qq_21057881/article/details/51034348

本文探讨了一个涉及蛋糕堆叠的问题，通过使用线段树维护动态规划来解决体积最大堆叠的问题。详细介绍了问题背景、算法思路以及具体实现过程。

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题意：有n个蛋糕，然后第i个蛋糕只能放在地上或者放在体积和编号都比他小的上面，然后问你体积最多能堆多大？

思路：用线段树维护DP，显然这个东西和lis（最长上升子序列）有一点像我们首先把每个东西的体积都离散化一下，然后我们选取比他小的最大值，然后进行更新就好了

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
typedef double SgTreeDataType;
struct treenode
{
  int L , R  ;
  double sum;
  int num;

};

treenode tree[500000];


inline void push_up(int o)
{
    tree[o].sum = max(tree[2*o].sum , tree[2*o+1].sum);
    if(tree[2*o].sum>tree[2*o+1].sum)
        tree[o].num=tree[2*o].num;
    else
        tree[o].num=tree[2*o+1].num;
}

inline void build_tree(int L , int R , int o)
{
    tree[o].L = L , tree[o].R = R,tree[o].sum = 0;
    tree[o].num = L;
    if (R > L)
    {
        int mid = (L+R) >> 1;
        build_tree(L,mid,o*2);
        build_tree(mid+1,R,o*2+1);
    }
}
inline void updata2(int QL,int QR,SgTreeDataType v,int o)
{
    int L = tree[o].L , R = tree[o].R;
    if (QL <= L && R <= QR)
    {
        tree[o].sum=max(tree[o].sum,v);
    }
    else
    {
        int mid = (L+R)>>1;
        if (QL <= mid) updata2(QL,QR,v,o*2);
        if (QR >  mid) updata2(QL,QR,v,o*2+1);
        push_up(o);
    }
}
inline SgTreeDataType query(int QL,int QR,int o)
{
    if(QR<QL)return 0;
    int L = tree[o].L , R = tree[o].R;
    if (QL <= L && R <= QR) return tree[o].sum;
    else
    {
        int mid = (L+R)>>1;
        SgTreeDataType res = 0;
        if (QL <= mid) res = max(query(QL,QR,2*o),res);
        if (QR > mid) res = max(query(QL,QR,2*o+1),res);
        push_up(o);
        return res;
    }
}


double h[maxn],r[maxn],v[maxn];
const double pi = acos(-1.0);
vector<double>V;
map<double,int> H;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%lf",&r[i],&h[i]),v[i]=pi*r[i]*r[i]*h[i];
        V.push_back(v[i]);
    }
    sort(V.begin(),V.end());
    V.erase(unique(V.begin(),V.end()),V.end());
    for(int i=0;i<V.size();i++)
        H[V[i]]=i+1;
    build_tree(1,n,1);
    for(int i=1;i<=n;i++)
    {
        double p = v[i]+query(1,H[v[i]]-1,1);
        updata2(H[v[i]],H[v[i]],p,1);
    }
    printf("%.12f\n",tree[1].sum);
    return 0;
}

Description

As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake.

Simple cake is a cylinder of some radius and height. The volume of the simple cake is equal to the volume of corresponding cylinder. Babaei has n simple cakes and he is going to make a special cake placing some cylinders on each other.

However, there are some additional culinary restrictions. The cakes are numbered in such a way that the cake number i can be placed only on the table or on some cake number j where j < i. Moreover, in order to impress friends Babaei will put the cake i on top of the cake j only if the volume of the cake i is strictly greater than the volume of the cake j.

Babaei wants to prepare a birthday cake that has a maximum possible total volume. Help him find this value.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of simple cakes Babaei has.

Each of the following n lines contains two integers r_i and h_i (1 ≤ r_i, h_i ≤ 10 000), giving the radius and height of the i-th cake.

Output

Print the maximum volume of the cake that Babaei can make. Your answer will be considered correct if its absolute or relative error does not exceed 10^- 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if $\text{[math]}$ .