2017/finals/strolls.cpp

// Hacker Cup 2017
// Final Round
// Fox Strolls
// Jacob Plachta

#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;

#define LL long long
#define LD long double
#define PR pair<int,int>

#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second

template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x*x); }

const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);

bool Read(int &x)
{
	char c,r=0,n=0;
	x=0;
		for(;;)
		{
			c=GETCHAR();
				if ((c<0) && (!r))
					return(0);
				if ((c=='-') && (!r))
					n=1;
				else
				if ((c>='0') && (c<='9'))
					x=x*10+c-'0',r=1;
				else
				if (r)
					break;
		}
		if (n)
			x=-x;
	return(1);
}

#define LIM 500000

int main()
{
	// vars
	int T,t;
	int N;
	int i,j,a,b,h;
	LL ans1,ans2;
	static int H[LIM];
	static PR P[LIM];
	stack<PR> ST;
	set<int> S;
	set<int>::iterator I;
	// testcase loop
	Read(T);
		Fox1(t,T)
		{
			// input
			Read(N);
				Fox(i,N)
				{
					Read(H[i]);
					P[i]=mp(H[i],i);
				}
			// PART 1: sum of distances
			ans1=0;
			// get base sum assuming no bridges
				Fox(i,N)
					ans1+=H[i];
			ans1*=N-1;
				Fox1(i,N-1)
					ans1+=(LL)i*(N-i);
			// process trees bottom-to-top
			sort(P,P+N);
			S.clear();
			S.insert(-1);
			S.insert(N);
				Fox(i,N)
				{
					h=P[i].x;
					j=P[i].y;
					// find interval of paths having this as their shortest tree
					I=S.lower_bound(j);
					b=*I - 1;
					I--;
					a=*I + 1;
					// subtract total distance saved by bridges
					ans1-=(LL)h*2*((LL)(j-a)*(b-j) + (b-a));
					// insert this tree
					S.insert(j);
				}
			// PART 2: number of bridges
			ans2=0;
			// process trees left-to-right
				while (!ST.empty())
					ST.pop();
			ST.push(mp(0,0));
				Fox1(i,N-1)
					if (H[i]>=H[i-1])
					{
						// at least as high as envelope's highest point, so build single bridge backwards
							if (ST.top().x==H[i-1])
								ST.pop();
						ST.push(mp(H[i-1],i));
						ans2++;
					}
					else
					{
						// clip off higher parts of envelope
						j=i-1;
							while (ST.top().x>H[i])
								j=ST.top().y,ST.pop();
						// build row of bridges back across, and update envelope
							if (!ST.top().x)
								ans2+=i-j;
							else
							{
								ans2+=i-ST.top().y;
								h=ST.top().x,ST.pop();
									if (H[i]!=h)
										ST.push(mp(h,j));
							}
						ST.push(mp(H[i],i));
					}
			// output
			printf("Case #%d: %lld %lld\n",t,ans1,ans2);
		}
	return(0);
}