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

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 .

Sample Input

Input

2
100 30
40 10

Output

942477.796077000

Input

4
1 1
9 7
1 4
10 7

Output

3983.539484752

Hint

In first sample, the optimal way is to choose the cake number 1.

In second sample, the way to get the maximum volume is to use cakes with indices 1, 2 and 4.