Convex hull trick/acquire.cpp

/*
ID: brian_bi21
PROG: acquire
LANG: C++
*/
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int pointer; //Keeps track of the best line from previous query
vector<long long> M; //Holds the slopes of the lines in the envelope
vector<long long> B; //Holds the y-intercepts of the lines in the envelope
//Returns true if either line l1 or line l3 is always better than line l2
bool bad(int l1,int l2,int l3)
{
	/*
	intersection(l1,l2) has x-coordinate (b1-b2)/(m2-m1)
	intersection(l1,l3) has x-coordinate (b1-b3)/(m3-m1)
	set the former greater than the latter, and cross-multiply to
	eliminate division
	*/
	return (B[l3]-B[l1])*(M[l1]-M[l2])<(B[l2]-B[l1])*(M[l1]-M[l3]);
}
//Adds a new line (with lowest slope) to the structure
void add(long long m,long long b)
{
	//First, let's add it to the end
	M.push_back(m);
	B.push_back(b);
	//If the penultimate is now made irrelevant between the antepenultimate
	//and the ultimate, remove it. Repeat as many times as necessary
	while (M.size()>=3&&bad(M.size()-3,M.size()-2,M.size()-1))
	{
		M.erase(M.end()-2);
		B.erase(B.end()-2);
	}
}
//Returns the minimum y-coordinate of any intersection between a given vertical
//line and the lower envelope
long long query(long long x)
{
	//If we removed what was the best line for the previous query, then the
	//newly inserted line is now the best for that query
	if (pointer>=M.size())
		pointer=M.size()-1;
	//Any better line must be to the right, since query values are
	//non-decreasing
	while (pointer<M.size()-1&&
	  M[pointer+1]*x+B[pointer+1]<M[pointer]*x+B[pointer])
		pointer++;
	return M[pointer]*x+B[pointer];
}
int main()
{
	int M,N,i;
	pair<int,int> a[50000];
	pair<int,int> rect[50000];
	freopen("acquire.in","r",stdin);
	freopen("acquire.out","w",stdout);
	scanf("%d",&M);
	for (i=0; i<M; i++)
		scanf("%d %d",&a[i].first,&a[i].second);
	//Sort first by height and then by width (arbitrary labels)
	sort(a,a+M);
	for (i=0,N=0; i<M; i++)
	{
		/*
		When we add a higher rectangle, any rectangles that are also
		equally thin or thinner become irrelevant, as they are
		completely contained within the higher one; remove as many
		as necessary
		*/
		while (N>0&&rect[N-1].second<=a[i].second)
			N--;
		rect[N++]=a[i]; //add the new rectangle
	}
	long long cost;
	add(rect[0].second,0);
	//initially, the best line could be any of the lines in the envelope,
	//that is, any line with index 0 or greater, so set pointer=0
	pointer=0;
	for (i=0; i<N; i++) //discussed in article
	{
		cost=query(rect[i].first);
		if (i<N)
			add(rect[i+1].second,cost);
	}
	printf("%lld\n",cost);
	return 0;
}