Submission #467387

#	Time	Username	Problem	Language	Result	Execution time	Memory
467387		ChrisGe123	Arranging Shoes (IOI19_shoes)	C++14	100 / 100	90 ms	16380 KiB

shoes

#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>;

const int maxn = 100002;
#define mp make_pair
#define pb push_back
#define f first
#define s second

const int mod = 1e9+7;

typedef pair<int, int> pii;
int n;

// for each shoe, we need the closest shoe to the right that is the opposite pair to it
int r[2*maxn];
vector<int> eachsize[2*maxn]; //eachsize[i] contains the positions of the shoes of size i-n in order
int sizepos[2*maxn]; //sizepos[i] tells us how many shoes of size i-n we've seen 
int fentree[2*maxn];

/*
We need to figure out using some PURS thing how to get the number of elements between i and r[i] that have already been taken care of
we can do this by just making a BIT where initially everything is 1, and then if the element i has been used, we make BIT[i] = 0


*/
int lsb(int a)
{
	return a & -a;
}
void change(int a, int c)
{
	while (a <= 2*n)
	{
		fentree[a] += c;
		a += lsb(a);
	}
}
ll sum(int a)
{
	ll s = 0;
	while (a > 0)
	{
		s += fentree[a];
		a -= lsb(a);
	}
	return s;
}

ll count_swaps(vector<int> shoes)
{
	n = shoes.size()/2;
	shoes.insert(shoes.begin(), 0);
	for (int i=1; i<=2*n; i++)
	{
		eachsize[shoes[i]+n].pb(i);
	}
	for (int i=1; i<=2*n; i++)
	{
		if (sizepos[shoes[i]+n] < (int) eachsize[-shoes[i]+n].size())
		{
			int temp = eachsize[-shoes[i]+n][sizepos[shoes[i]+n]];
			if (temp > i)
			{
				r[i] = temp;
			}
		}
		sizepos[shoes[i]+n]++;
	}

	ll ans = 0;
	for (int i=1; i<=2*n; i++)
	{
		change(i, 1);
	}
	for (int i=1; i<=2*n; i++)
	{
		if (r[i] != 0)
		{
			if (r[i] != i+1)
			{
				ans += sum(r[i]-1) - sum(i);
				//cerr << sum(r[i]-1) << " " << sum(i) << endl;
			}
			if (shoes[i] > 0)
			{
				ans++;
			}
			change(r[i], -1);
		}
	}
	return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 15 / 15