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
942477.796077000
4 1 1 9 7 1 4 10 7
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.
题意：找到累计最多的蛋糕体积。两个蛋糕累计条件（j < i && v[j] < v[i]）。
思路：可以用最长上升子序列的 dp 来做，体积为数列元素。 但是数据太大，查询 j = 1 到 j < i 时需要浪费 O(n)的时间。时间复杂度就是 O(n^2) 肯定会TEL。因此查询以及更新利用线段树来维护。将区间的从小到大抽象成体积 v[i] 从小到大，每次利用线段树查询和更新 dp[i]即可。时间复杂度为 O(n*logn)。
代码：
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cmath> 5 using namespace std; 6 7 struct Node{ 8 int l,r; 9 long long Max; 10 }node[100005*4]; 11 long long v[100005]; 12 long long x[100005]; 13 long long dp[100005]; 14 const double PI = acos(-1.0); 15 16 void build(int id,int l,int r){ 17 int mid = (l+r)>>1; 18 node[id].l = l; 19 node[id].r = r; 20 if(l == r){ 21 node[id].Max = 0; 22 return ; 23 } 24 build(id*2,l,mid); 25 build(id*2+1,mid+1,r); 26 27 node[id].Max = max(node[id*2].Max,node[id*2+1].Max); 28 } 29 30 void update(int id,int k,long long p){ //更新 k 处的值为 p 31 int ll = node[id].l; 32 int rr = node[id].r; 33 int mid = (ll+rr)>>1; 34 35 if(ll == rr){ 36 node[id].Max = p; 37 return ; 38 } 39 40 if(k <= mid) update(id*2,k,p); 41 else 42 update(id*2+1,k,p); 43 node[id].Max = max(node[id*2].Max,node[id*2+1].Max); 44 } 45 46 long long query(int id,int l,int r){ 47 int ll = node[id].l; 48 int rr = node[id].r; 49 int mid = (ll+rr)>>1; 50 51 if(l > r) return 0; 52 53 if(ll== l && rr == r) 54 return node[id].Max; 55 else if(r <= mid) 56 return query(id*2,l,r); 57 else if(l > mid) 58 return query(id*2+1,l,r); 59 else 60 return max(query(id*2,l,mid),query(id*2+1,mid+1,r)); 61 } 62 int main() 63 { 64 int n; 65 while(cin>>n){ 66 long long r,h; 67 for(int i = 1;i <= n;i ++){ 68 scanf("%I64d%I64d",&r,&h); 69 v[i] = r*r*h; 70 dp[i] = v[i]; 71 x[i] = v[i]; 72 } 73 sort(x+1,x+n+1); 74 long long ans = 0; 75 build(1,1,n); 76 for(int i = 1;i <= n;i ++){ 77 int cur = lower_bound(x+1,x+n+1,v[i]) - x; //前闭后开区间找大于或等于 v[i]的第一个位置。 78 //wa了好久都不知道为啥。 79 dp[i] = max(dp[i],query(1,1,cur-1)+v[i]); 80 update(1,cur,dp[i]); 81 ans = max(ans,dp[i]); 82 } 83 printf("%.15f ",ans*PI); 84 } 85 return 0; 86 }