Codeforces Round #343 (Div. 2) D. Babaei and Birthday Cake （复杂的DP）

题目链接：http://codeforces.com/problemset/problem/629/D

D. Babaei and Birthday Cake time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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.

题意：给你n个圆柱体的底面半径r和高h，求最大能放在一起的总体积，条件是i放到j上 Vi >= Vj  并且i > j

解析：显然是一个DP题，但数据比较大，需要优化，这里用到了map优化，就是mp[i]一直是最优的，每输入一个圆柱体i，就要二分mp里面的>=该体积的值j，存下当前值与二分到的j的上一个相加，cnt = (--R)->second + v; 然后遍历把小于cnt的值都给删了，再把mp[v] = cnt;存进去，这样就保证了所有体积都最优

借鉴 hpu 纯真学姐 博客 ：http://blog.csdn.net/zwj1452267376/article/details/50720228

代码：

#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<map> #include<cmath> #include<string> #define N 1009 using namespace std; const int INF = 0x3f3f3f3f; typedef __int64 LL; map<LL, LL> mp; int main() { int n; LL r, h; scanf("%d", &n); mp[0] = 0; for(int i = 1; i <= n; i++) { scanf("%I64d %I64d", &r, &h); LL v = r * r * h; map<LL, LL>::iterator L = mp.lower_bound(v); map<LL, LL>::iterator R = L; LL cnt = (--R)->second + v; R = L; for(; R != mp.end() && R->second <= cnt; R++) ; mp.erase(L, R); //删除[l,r)之间元素 mp[v] = cnt; } printf("%.9f\n", (--mp.end())->second * acos(-1.0)); return 0; }