D. Babaei and Birthday Cake
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 .
Examples
Input
2
100 30
40 10
Output
942477.796077000
Input
4
1 1
9 7
1 4
10 7
Output
3983.539484752
Note
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.
對於當前的第i個，肯定是找到容量比他小的中能得到的總容量最大的那個放上去，所以離散化一下，線段樹保存容量排序後第i個能得到的最佳答案，dp[i]表示以第i個結尾能得到的最佳答案。
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100010;
const int maxm=1010;
const int MOD=1e9+7;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
int N,nx;
LL X[maxn];
struct Node{
int r,h;
}cyl[maxn];
LL dp[maxn];
LL ans;
struct IntervalTree{
LL maxv[maxn<<2];
void build(int o,int l,int r){
maxv[o]=0;
if(l==r)return ;
int mid=(l+r)>>1;
build(o<<1,l,mid);
build(o<<1|1,mid+1,r);
}
void pushup(int o){
maxv[o]=max(maxv[o<<1|1],maxv[o<<1]);
}
void query(int o,int l,int r,int q1,int q2){
if(l>=q1&&q2>=r){
ans=max(ans,maxv[o]);
return ;
}
int mid=(l+r)>>1;
if(q1<=mid)query(o<<1,l,mid,q1,q2);
if(q2>mid) query(o<<1|1,mid+1,r,q1,q2);
}
void update(int o,int l,int r,int pos,LL val){
if(l==r){
maxv[o]=val;
return;
}
int mid=(l+r)>>1;
if(pos<=mid)update(o<<1,l,mid,pos,val);
else update(o<<1|1,mid+1,r,pos,val);
pushup(o);
}
}tree;
int main(){
while(scanf("%d",&N)!=EOF){
for(int i=1;i<=N;i++){
scanf("%d%d",&cyl[i].r,&cyl[i].h);
X[i]=1LL*cyl[i].r*cyl[i].r*cyl[i].h;
}
sort(X+1,X+1+N);
nx=unique(X+1,X+1+N)-X-1;
tree.build(1,1,nx);
for(int i=1;i<=N;i++){
LL vol=1LL*cyl[i].r*cyl[i].r*cyl[i].h;
int pos=lower_bound(X+1,X+nx+1,vol)-X;
if(pos-1==0){
dp[i]=vol;
} else {
ans=0;
tree.query(1,1,nx,1,pos-1);
dp[i]=ans+vol;
}
tree.update(1,1,nx,pos,dp[i]);
}
ans=0;
for(int i=1;i<=N;i++){
ans=max(ans,dp[i]);
}
printf("%.9f\n",1.0*PI*ans);
}
return 0;
}