Submission #1204031

#	Time	Username	Problem	Language	Result	Execution time	Memory
1204031		countless	Arranging Shoes (IOI19_shoes)	C++20	100 / 100	222 ms	30296 KiB

shoes

#include "shoes.h"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

const ll MOD = 998244353;
const ll INF = 1e18;
const ld EPS = 1e-12;

#define endl "\n"
#define sp <<" "<<
#define REP(i, a, b) for(ll i = a; i < b; i++)
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())

struct custom_hash {
	static uint64_t splitmix64(uint64_t x) {
		// http://xorshift.di.unimi.it/splitmix64.c
		x += 0x9e3779b97f4a7c15;
		x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
		x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
		return x ^ (x >> 31);
	}

	size_t operator()(uint64_t x) const {
		static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
		return splitmix64(x + FIXED_RANDOM);
	}
};

template <typename Key, typename Value>
using hash_map = unordered_map<Key, Value, custom_hash>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// uniform_int_distribution<int>(a, b)(rng);
// shuffle(all(a), rng);

// https://www.geeksforgeeks.org/inversion-count-in-array-using-merge-sort/#expected-approach-using-merge-sort-onlog-n-time-and-on-space
// This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
// and also counts inversions in the whole subarray arr[l..r]
ll countAndMerge(vector<int>& arr, int l, int m, int r) {
  
    // Counts in two subarrays
    int n1 = m - l + 1, n2 = r - m;

    // Set up two vectors for left and right halves
    vector<int> left(n1), right(n2);
    for (int i = 0; i < n1; i++)
        left[i] = arr[i + l];
    for (int j = 0; j < n2; j++)
        right[j] = arr[m + 1 + j];

    // Initialize inversion count (or result) and merge two halves
    ll res = 0;
    int i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {

        // No increment in inversion count if left[] has a 
        // smaller or equal element
        if (left[i] <= right[j]) 
            arr[k++] = left[i++];
      
        // If right is smaller, then it is smaller than n1-i 
      	// elements because left[] is sorted
        else {
            arr[k++] = right[j++];
            res += (n1 - i);
        }
    }

    // Merge remaining elements
    while (i < n1)
        arr[k++] = left[i++];
    while (j < n2)
        arr[k++] = right[j++];

    return res;
}

// Function to count inversions in the array
ll countInv(vector<int>& arr, int l, int r){
    ll res = 0;
    if (l < r) {
        int m = (r + l) / 2;

        // Recursively count inversions in the left and 
        // right halves
        res += countInv(arr, l, m);
        res += countInv(arr, m + 1, r);

        // Count inversions such that greater element is in 
      	// the left half and smaller in the right half
        res += countAndMerge(arr, l, m, r);
    }
    return res;
}

ll inversionCount(vector<int> &arr) {
  	int n = arr.size();
  	return countInv(arr, 0, n-1);
}

ll count_swaps(vector<int> s) {
	int n = s.size();
	// 2, 1, -1, -2
	// 1, 2, -2, -1
	map<int, vector<int>> pos;
	REP(i, 0, n) {
		pos[s[i]].push_back(i);
	}

	int id = 1;
	for (auto &x : pos) {
		if (x.first > 0) break;

		int m = x.second.size();
		REP(i, 0, m) {
			s[x.second[i]] = -id;
			s[pos[-x.first][i]] = id;
			id++;
		}
	}

	vector<int> dup(n);
	map<int, int> have;
	int curr = 1;
	REP(i, 0, n) {
		if (have[abs(s[i])] != 0) {
			dup[i] = have[abs(s[i])] + (s[i] > 0);
			have[abs(s[i])] = 0;
		} else {
			// cerr << i << endl;
			dup[i] = curr + (s[i] > 0);
			have[abs(s[i])] = curr;
			curr += 2;
		}
	}

	// REP(i, 0, n) {
	// 	cerr << dup[i] << " ";
	// }	cerr << endl;

	// return inversionCount(dup);
	ll ans = inversionCount(dup);
	// assert(ans >= 0 and ans <= (n * (n+1)) / 2);
	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