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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#549537#860. We apologize for any inconvenienceRafat_KabirRE 0ms0kbC++204.6kb2024-09-06 17:13:362024-09-06 17:13:37

Judging History

你现在查看的是最新测评结果

  • [2024-09-06 17:13:37]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2024-09-06 17:13:36]
  • 提交

answer

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <time.h>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>

using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <iostream>

using namespace __gnu_pbds;
using namespace std;
template <class T>
using Tree =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// to erase in multiset-> less_equal<T> and 
// s.erase(s.find_by_order(s.order_of_key(x)))
// lower_bound(x)=>(cannot use the stl lower_bound function)
// ll idx = s.order_of_key(x)
// if(idx == s.size()) -> no lower_bound
// else lb = *s.find_by_order(idx) // as 0-indexing
// idx-1 will give highest value which is strictly less than x
// for upper_bound->do the same with (x+1)

typedef long long ll;
typedef long double ld;
typedef pair<int,int> p32;
typedef pair<ll,ll> p64;
typedef tuple<ll, ll, ll> t64;
typedef vector<t64> vt64;
typedef vector<vt64> vvt64;
typedef pair<double,double> pdd;
typedef vector<ll> v64;
typedef vector<int> v32;
typedef vector<vector<int> > vv32;
typedef vector<vector<ll> > vv64;
typedef vector<vector<p64> > vvp64;
typedef vector<p64> vp64;
typedef vector<p32> vp32;
typedef vector<vector<p32> > vvp32;
typedef vector<bool> vb;
ll mod =  1e9+7, MOD = 998244353;
double eps = 1e-12;
// #define forn(i,e) for(ll i = 0; i < e; i++)
#define FOR(s, e, i) for(int i = s; i <= e; i++)
// #define rforn(i,s) for(ll i = s; i >= 0; i--)
#define ROF(s ,e, i) for(int i = s; i >= e; i--)
#define coutAll(A) for(auto asdafas : A) cout <<  asdafas << " "; cout << "\n";
#define foutAll(A) for(auto asdafas : A) fout <<  asdafas << " "; cout << "\n";
#define cinAll(A) for(auto &asdafas : A) cin >>  asdafas;
#define finAll(A) for(auto &asdafas : A) fin >>  asdafas;
#define minpq priority_queue<ll, v64, greater<ll>>
#define maxpq priority_queue<ll> 
#define ln "\n"
#define dbg(x) cout<<#x<<" = "<<x<<ln
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define fi first
#define se second
ll inf = LLONG_MAX;
#define fast_cin() ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())
#define yes cout<<"YES\n"
#define no cout<<"NO\n"
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef pair<ll, ll> pll;
typedef pair<ll, ll> pii;
//#define MAXN 1000000

void solve(int it)
{
    int n, k;
    scanf("%d%d", &n, &k);
    vv64 dp(n+k, v64(n+k, inf));
    FOR(0, n - 1, i) dp[i][i] = 0;
    FOR(0, k - 1, i){
        int r; cin >> r;
        FOR(0, r - 1, j){
            ll x;cin >> x;
            --x;
            dp[x][i+n] = dp[i+n][x] = 1;
        }
        // FOR(0, i-1, j){
        //     dp[i+n][j+n] = dp[j+n][i+n] = 1;
        // }
    }
    int s; 
    scanf("%d", &s);
    v32 pos(k,-1);
    FOR(0, s - 1, i){
        ll id;
        scanf("%d", &id);
        --id;
        pos[id] = s-i;
    }
    v32 A(n+k);
    iota(all(A), 0);
    // coutAll(pos);
    sort(A.begin()+n, A.begin()+n+k, [&](int a, int b){
        return pos[a-n] < pos[b-n];
    });
    // FOR(n, n+k-1, i) cout << A[i]+1-n << " ";
    // cout << "\n";
    // coutAll(A);
    v64 ans(s+1, 0);
    // cout << "done\n";
    FOR(0, n+k-1, l){
        ll mx = 0;
        if(A[l]-n >= 0 && pos[A[l]-n] != -1){
            // cout << s+1-pos[A[l]-n] << "->";
            FOR(0, n-1, i){
                FOR(0, n - 1, j){
                    if(dp[i][j] == inf) continue;
                    if(dp[i][j] > 0)
                        mx = max(mx, dp[i][j]/2-1);
                }
            }
            ans[s+1-pos[A[l]-n]] = mx;
        }
        FOR(0, n+k-1, i){
            FOR(0, n+k-1, j){
                if(dp[A[i]][A[l]] == inf) continue;
                if(dp[A[l]][A[j]] == inf) continue;
                if(dp[A[i]][A[j]] <= dp[A[i]][A[l]] + dp[A[l]][A[j]]) continue;
                dp[A[i]][A[j]] = dp[A[i]][A[l]] + dp[A[l]][A[j]];
            }
        }
    }
    // cout <<"\n";
    // coutAll(ans);
    FOR(0, n-1, i){
        FOR(0, n - 1, j){
            if(dp[i][j] == inf) continue;
            ans[0] = max(ans[0], dp[i][j]/2-1);
        }
    }
    for(auto x : ans) cout << x << "\n";
}


int main()
{
    // fast_cin();    
    ll t = 1;
    cin >> t;
    for(int it=1; it<=t; it++)
    {
        //cout << "Case " << it << ": ";
        solve(it);
    }
    return 0;
}


詳細信息

Test #1:

score: 0
Runtime Error

input:

1
5 4
3 1 3 5
2 1 4
2 2 3
2 2 4
3
1
4
3

output:


result: