Submission #331050

#	Time	Username	Problem	Language	Result	Execution time	Memory
331050		ChrisGe123	Palembang Bridges (APIO15_bridge)	C++14	0 / 100	3 ms	364 KiB

bridge

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <class T> using Tree = tree<T, null_type, less<T>,
    rb_tree_tag, tree_order_statistics_node_update>;

//g++ -std=c++17 -O2 -fsanitize=undefined /usr/local/Cellar/gcc/10.2.0/include/c++/10.2.0/x86_64-apple-darwin19/bits/stdc++.h
#define maxn 100005
#define mod 1000000007
int n, k;
pair<ll, pair<ll, ll> > interval[maxn]; 
ll ans;


Tree<pair<ll, int> > lendpoints;
Tree<pair<ll, int> > rendpoints;

void setupIO(string s)
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	freopen((s + ".in").c_str(), "r", stdin);
	freopen((s + ".out").c_str(), "w", stdout);
}

ll f(int i, ll bridge)
{
	ll l = interval[i].second.first; ll r = interval[i].second.second;
	if (l <= bridge && bridge <= r)
	{
		return 0;
	}
	else if (bridge < l)
	{
		return 2 * (l-bridge);
	}
	else
	{
		return 2 * (bridge-r);
	}
}
int main()
{
	setupIO("bridges");
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> k >> n;
	int counter = 0;
	ll pre = 0;
	lendpoints.clear();
	rendpoints.clear();
	for (int i=0; i<n; i++)
	{
		char p, q;
		ll s, t;
		cin >> p >> s >> q >> t;
		if (p == q)
		{
			pre += abs(s - t);
		}
		else
		{
			ll l = min(s, t);
			ll r = max(s, t);
			interval[counter++] = make_pair(s+t, make_pair(l, r));
			pre += r - l + 1;
		}
	}
	//use the below version to make it easier to randomly generate test cases 
	// for (int i=0; i<n; i++)
	// {
	// 	ll s, t;
	// 	cin >> s >> t;
	// 	ll l = min(s, t);
	// 	ll r = max(s, t);
	// 	interval[counter++] = make_pair(s+t, make_pair(l, r));
	// 	pre += r - l + 1;
	// }
	n = counter;
	ans = 0;
	//we just want to find the median of the endpoints 
	sort(interval, interval+n);
	for (int i=0; i<n; i++)
	{
		rendpoints.insert(make_pair(interval[i].second.first, 2*i));
		rendpoints.insert(make_pair(interval[i].second.second, 2*i+1));
	}
	ll xone = rendpoints.find_by_order(n-1)->first;
	//cerr << x << endl;
	for (int i=0; i<n; i++)
	{
		ll l = interval[i].second.first; ll r = interval[i].second.second;
		if (l <= xone && xone <= r)
		{
			continue;
		}
		else if (xone < l)
		{
			ans += 2 * (l-xone);
		}
		else
		{
			ans += 2 * (xone-r);
		} 
	}
	if (k == 2)
	{
		 //sorted in order of increasing midpoint
		// cerr << ans << endl;
		// cerr << endl;
		// cerr << endl;
		ll temp = ans;
		for (int i=0; i<n-1; i++)
		{
			//things 0 to i take the first bridge, i+1 to n-1 take the second bridge
			lendpoints.insert(make_pair(interval[i].second.first, 2*i));
			lendpoints.insert(make_pair(interval[i].second.second, 2*i+1));
			ll x = lendpoints.find_by_order(i)->first;

			rendpoints.erase(make_pair(interval[i].second.first, 2*i));
			rendpoints.erase(make_pair(interval[i].second.second, 2*i+1));
			ll y = rendpoints.find_by_order(n-i-2)->first;
			temp += f(i,x) -f(i, y);
		
			/* 
			To figure out the change from x_prev to x for items 1 to i-1, 

			Claim: each new point added has its midpoint to the right of the current median
				proof is by induction and using the fact that sorted order of midpoints
				in particular, this means that either one point left of median and one right or two right, so the median never moves back
			If the median stays the same, no change
			If the median moves forward by one point
				if it moved to the left endpoint of point i: no change
				if it moved to any other point: it just moved to prevhighx from prevlowx, so no change
			NO CHANGE FOR ITEMS 1 TO i-1

			To figure out the change for item i: just calculate directly f(i, x) - f(i, oldy)

			To figure out the change for items i+1 to n-1: 
				goes down by f(i, newy) - f(i, oldy): let's see why
					if i was both left of old median and both left of new median: sum wouldn't change if you moved from old to new, but now that we don't factor in the f(i, new) - f(i, old), we have to subtract that
					if i was one left one right of old median and both left of new median: same as above except f(i, old) = 0
					if i was one left one right of old median and one left one right of new median: no change at all, this works

			*/
			// cerr << i << endl;
			// cerr << temp << endl;
			// cerr << endl;
			ans = min(ans, temp);
		}
	}
	cout << ans + pre << endl;
}

Compilation message (stderr)

bridge.cpp: In function 'void setupIO(std::string)':
bridge.cpp:25:9: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   25 |  freopen((s + ".in").c_str(), "r", stdin);
      |  ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
bridge.cpp:26:9: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
   26 |  freopen((s + ".out").c_str(), "w", stdout);
      |  ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 0 / 8
• Subtask 2: 0 / 14
• Subtask 3: 0 / 9
• Subtask 4: 0 / 32
• Subtask 5: 0 / 37