Submission #298804

#	Time	Username	Problem	Language	Result	Execution time	Memory
298804		square1001	Arranging Shoes (IOI19_shoes)	C++14	65 / 100	571 ms	40184 KiB

shoes

#include "shoes.h"
#include <map>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const int inf = 1012345678;
long long count_inversions(int n, vector<int> a, vector<int> b) {
	map<int, int> da, db;
	vector<int> aprecnt(n);
	for(int i = 0; i < n; ++i) {
		aprecnt[i] = da[a[i]]++;
	}
	map<pair<int, int>, int> dats;
	for(int i = 0; i < n; ++i) {
		dats[make_pair(a[i], aprecnt[i])] = i;
	}
	vector<int> perm(n);
	for(int i = 0; i < n; ++i) {
		int bprecnt = db[b[i]]++;
		perm[i] = dats[make_pair(b[i], bprecnt)];
	}
	vector<int> bit(n + 1);
	vector<int> iperm(n);
	for(int i = 0; i < n; ++i) {
		iperm[perm[i]] = i;
	}
	long long ans = 0;
	for(int i = n - 1; i >= 0; --i) {
		int pos = iperm[i];
		for(int j = pos; j >= 1; j -= j & (-j)) {
			ans += bit[j];
		}
		for(int j = pos + 1; j <= n; j += j & (-j)) {
			++bit[j];
		}
	}
	return ans;
}
long long count_swaps(std::vector<int> s) {
	int N = s.size() / 2;
	bool subtask3 = true, subtask4 = true;
	for(int i = 1; i < 2 * N; ++i) {
		if(abs(s[i]) != abs(s[i - 1])) {
			subtask3 = false;
		}
	}
	for(int i = 0; i < N; ++i) {
		if(!(s[i] < 0 && s[i + N] > 0 && s[i] + s[i + N] == 0)) {
			subtask4 = false;
		}
	}
	if(subtask3) {
		vector<int> seq(2 * N);
		for(int i = 0; i < N; ++i) {
			seq[i * 2] = -abs(s[0]);
			seq[i * 2 + 1] = abs(s[0]);
		}
		long long ans = count_inversions(2 * N, seq, s);
		return ans;
	}
	else if(subtask4) {
		vector<int> seq(2 * N);
		for(int i = 0; i < N; ++i) {
			seq[i * 2] = s[i];
			seq[i * 2 + 1] = -s[i];
		}
		long long ans = count_inversions(2 * N, seq, s);
		return ans;
	}
	else if(N <= 8) {
		// subtask 1, 2
		vector<int> sizes;
		for(int i = 0; i < 2 * N; ++i) {
			if(s[i] > 0) {
				sizes.push_back(s[i]);
			}
		}
		vector<int> perm(N);
		for(int i = 0; i < N; ++i) perm[i] = i;
		int ans = inf;
		do {
			vector<int> seq(2 * N);
			for(int i = 0; i < N; ++i) {
				seq[i * 2] = -sizes[perm[i]];
				seq[i * 2 + 1] = sizes[perm[i]];
			}
			int res = count_inversions(2 * N, seq, s);
			ans = min(ans, res);
		} while(next_permutation(perm.begin(), perm.end()));
		return ans;
	}
	return -1;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 15 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 15