제출 #358436

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
358436		Lam_lai_cuoc_doi	Xylophone (JOI18_xylophone)	C++17	0 / 100	51 ms	644 KiB

xylophone

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
 
#include "xylophone.h"
 
void solve(int n)
{
    vector<vector<int>> a(n + 1, vector<int>(n + 1));
    vector<set<pair<int, int>>> s(n + 1);
    for (int i = 1; i < n; ++i)
    {
        a[i][i + 1] = query(i, i + 1);
        s[a[i][i + 1]].insert({i, i + 1});
    }
    for (int i = 1; i < n - 1; ++i)
    {
        int v = query(i, i + 2);
        if (v == a[i][i + 1] + a[i + 1][i + 2])
        {
            a[i][i + 2] = v;
            s[v].insert({i, i + 2});
        }
        else
        {
            if (v == a[i][i + 1])
            {
                a[i][i + 2] = v - a[i + 1][i + 2];
                s[a[i][i + 2]].insert({i, i + 2});
            }
            else
            {
                a[i][i + 2] = v - a[i][i + 1];
                s[a[i][i + 2]].insert({i, i + 2});
            }
        }
    }
 
    for (int i = 3; i < n; ++i)
    {
        int sum(0);
        for (int j = 1; j < i - 1; ++j)
            sum += a[j][j + 1];
        for (int j = 1; j <= n - i; ++j)
        {
            sum -= a[j - 1][j];
            sum += a[j + i - 1][j + i];
            int v = query(j, i + j);
            if (v == sum)
            {
                a[j][i + j] = v;
                s[v].insert({j, i + j});
            }
            else
            {
                auto t = s[v].lower_bound({j, 0});
                if (t->first == j)
                {
                    a[j][i + j] = v - a[t->second][j + i];
                    s[a[j][i + j]].insert({j, i + j});
                }
                else if (t->second == j + i)
                {
                    a[j][i + j] = v - a[j][t->first];
                    s[a[j][j + i]].insert({j, j + i});
                }
                else
                {
                    a[j][i + j] = v - min(a[j][t->first] + a[t->second][j + i], a[j][t->second] + a[t->first][j + i]);
                    s[a[j][j + i]].insert({j, j + i});
                }
            }
        }
    }
 
    int pos(-1);
    for (int i = 1; i <= n; ++i)
    {
        for (int j = i + 1; j <= n; ++j)
            if (a[i][j] == n - 1)
            {
                pos = i;
                break;
            }
        if (pos != -1)
            break;
    }
    for (int i = 1; i <= n; ++i)
    {
        if (i == pos)
            answer(i, 1);
        else
            answer(i, 1 + (i < pos ? a[i][pos] : a[pos][i]));
    }
}
 

• Subtask 1: 0 / 11
• Subtask 2: 0 / 36
• Subtask 3: 0 / 53