题目连接：
http://www.codeforces.com/contest/629/problem/D
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 ri and hi (1 ≤ ri, hi ≤ 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
2
100 30
40 10
Sample Output
942477.796077000
Hint
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？
题解：
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
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
typedef double SgTreeDataType;
struct treenode
{
int L , R ;
double sum;
int num;
void updata(SgTreeDataType v)
{
sum += v;
}
};
treenode tree[500000];
inline void push_down(int o)
{
}
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
{
push_down(o);
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
{
push_down(o);
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;
}
