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.
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.
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 
.
2
100 30
40 10
942477.796077000
4
1 1
9 7
1 4
10 7
3983.539484752
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.
线段树维护比当前值a[i]小的最大dp[j]的值。j<i
我们可以用体积作为线段树的横坐标（就那个意思）.
也可以直接上。（就是官方题解）
/* *********************************************** Author :guanjun Created Time :2016/2/24 9:56:13 File Name :cf343d.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 101000 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; struct Node{ int x,y; }; struct cmp{ bool operator()(Node a,Node b){ if(a.x==b.x) return a.y> b.y; return a.x>b.x; } }; bool cmp(int a,int b){ return a>b; } struct node{ ll Max; int l,r; int getmid(){return (l+r)/2;} }nod[maxn*4]; void push_up(int i){ nod[i].Max=max(nod[i<<1].Max,nod[i<<1|1].Max); } void build(int i,int l,int r){ nod[i].l=l,nod[i].r=r; if(l==r){ nod[i].Max=0; return ; } int mid=(l+r)/2; build(i<<1,l,mid); build(i<<1|1,mid+1,r); } void update(int i,int k,ll w){ if(nod[i].l==k&&nod[i].r==k){ nod[i].Max=max(nod[i].Max,w);return ; } int mid=nod[i].getmid(); if(k<=mid)update(i<<1,k,w); else update(i<<1|1,k,w); push_up(i); } double query(int i,int l,int r){ if(r<l)return 0; if(nod[i].l==l&&nod[i].r==r){ return nod[i].Max; } int mid=nod[i].getmid(); if(r<=mid)return query(i<<1,l,r); else if(l>mid)return query(i<<1|1,l,r); else return max(query(i<<1,l,mid),query(i<<1|1,mid+1,r)); } map<ll,int>mp; ll a[maxn],b[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int n; while(cin>>n){ ll x,y; mp.clear(); for(int i=1;i<=n;i++){ cin>>x>>y; b[i]=a[i]=x*x*y; } sort(b+1,b+1+n); int num=0; for(int i=1;i<=n;i++){ if(!mp[b[i]])mp[b[i]]=++num; } build(1,1,n); //puts("YEs"); for(int i=1;i<=n;i++){ //cout<<mp[a[i]]<<endl; ll dp=query(1,1,mp[a[i]]-1)+a[i]; update(1,mp[a[i]],dp); } printf("%.8lf ",3.1415926*nod[1].Max); } return 0; }
/* *********************************************** Author :guanjun Created Time :2016/2/23 21:49:30 File Name :cf343d.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 0x3f3f3f3f #define maxn 101100 #define cle(a) memset(a,0,sizeof(a)) #define X first #define Y second const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; priority_queue<int,vector<int>,greater<int> >pq; pair<ll,int>a[maxn]; struct node{ int l,r; ll Max; }nod[maxn*4]; void push_up(int i){ nod[i].Max=max(nod[i<<1].Max,nod[i<<1|1].Max); } void build(int i,int l,int r){ nod[i].l=l; nod[i].r=r; if(l==r){ nod[i].Max=0; return ; } int mid=(l+r)/2; build(i<<1,l,mid); build(i<<1|1,mid+1,r); push_up(i); } void update(int i,int k,ll w){ if(nod[i].l==k&&nod[i].r==k){ nod[i].Max=max(nod[i].Max,w); return ; } int mid=(nod[i].l+nod[i].r)/2; if(k<=mid)update(i<<1,k,w); else update(i<<1|1,k,w); push_up(i); } double query(int i,int l,int r){ if(nod[i].l==l&&nod[i].r==r){ return nod[i].Max; } int mid=(nod[i].l+nod[i].r)/2; if(r<=mid)return query(i<<1,l,r); else if(l>mid)return query(i<<1|1,l,r); else return max(query(i<<1,l,mid),query(i<<1|1,mid+1,r)); } ll dp[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int n; while(cin>>n){ ll x,y; for(int i=1;i<=n;i++){ scanf("%I64d %I64d",&x,&y); a[i].X=x*x*y; a[i].Y=-i; } build(1,1,n); sort(a+1,a+1+n); for(int i=1;i<=n;i++){ int id=-a[i].Y; //cout<<id<<endl; dp[id]=query(1,1,id)+a[i].X; update(1,id,dp[id]); } printf("%.8lf ",nod[1].Max*3.1415926); } return 0; }
注意id 赋值为-i。这样才能保证相同的体积出现时我们取最后那一个。这里不用-i也行，也可以自己写一个比较函数。