add ACQUIRE

acquire.cpp

@@ -0,0 +1,85 @@
+// 2014-10-10
+// originally implemented c. Jan 2010
+#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];
+	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;
+}