# Convex hull trick/acquire.cpp

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/* ID: brian_bi21 PROG: acquire LANG: C++ */ /* 6th line from the end initially : if (i<N) now : if (i < N-1) by : pktiw */ #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-1) add(rect[i+1].second,cost); } printf("%lld\n",cost); return 0; }