Submission #1204030

#TimeUsernameProblemLanguageResultExecution timeMemory
1204030countlessArranging Shoes (IOI19_shoes)C++20
50 / 100
158 ms30152 KiB
#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; }
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