CF 319C(Kalila and Dimna in the Logging Industry-斜率DP,注意叉积LL溢出)

Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.

The manager of logging factory wants them to go to the jungle and cut n trees with heights a₁, a₂, ..., a_n. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height a_i in the beginning), then the cost of charging the chain saw would be b_i. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, a_i < a_j and b_i > b_j and also b_n = 0and a₁ = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.

They want you to help them! Will you?

Input

The first line of input contains an integer n (1 ≤ n ≤ 10⁵). The second line of input contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹). The third line of input contains n integers b₁, b₂, ..., b_n (0 ≤ b_i ≤ 10⁹).

It's guaranteed that a₁ = 1, b_n = 0, a₁ < a₂ < ... < a_n and b₁ > b₂ > ... > b_n.

Output

The only line of output must contain the minimum cost of cutting all the trees completely.

Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Sample test(s)
input
5
1 2 3 4 5
5 4 3 2 0
output
25
input
6
1 2 3 10 20 30
6 5 4 3 2 0
output
138
 
 
        

这题就是斜率DP(555...早知道直接做第3题)

方程我就不写了

long long 溢出我居然De了一上午Bug才发现，我……

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a))
#define MEMI(a) memset(a,127,sizeof(a))
#define MEMi(a) memset(a,128,sizeof(a))
#define MAXN (100000+10)
#define size (tail-head+1)
int n;
long long a[MAXN],b[MAXN],f[MAXN]={0};
struct node
{
	int i;
	long long x,y;
	node(){}
	node(int _i):i(_i),x(b[i]),y(f[i]){}
	friend long double k(node a,node b){return (long double)(a.y-b.y)/(long double)(a.x-b.x);	}
}q[MAXN];
int head=1,tail=1;
struct V
{
	long long x,y;
	V(node a,node b):x(b.x-a.x),y(b.y-a.y){}
	friend long long operator*(V a,V b){return ((double)a.x/a.y-(double)b.x/b.y)>0?1:-1;	}
};
int main()
{
//	freopen("CF319C.in","r",stdin);
//	freopen("CF319C.out","w",stdout);
	
	cin>>n;
	For(i,n) cin>>a[i];
	For(i,n) cin>>b[i];
	f[1]=b[1];
	q[1]=node(1);
	Fork(i,2,n)
	{
		while (size>=2&&k(q[head],q[head+1])>1-a[i] ) head++;
		int j=q[head].i;
		f[i]=f[j]+(long long)(a[i]-1)*(long long)b[j]+b[i];
		while (size>=2&&(V(q[tail-1],q[tail])*V(q[tail],node(i))>0)) tail--; //维护上凸壳 
		q[++tail]=node(i);
	}
	//For(i,n) cout<<b[i]<<' ';cout<<endl;
	//For(i,n) cout<<f[i]<<' ';cout<<endl;
	cout<<f[n]<<endl;
	
	return 0;
}