QOJ.ac

QOJ

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
#429069#6516. New but Nostalgic Problemdnialh#TL 0ms3512kbC++202.9kb2024-06-02 00:28:372024-06-02 00:28:38

Judging History

你现在查看的是最新测评结果

  • [2024-06-02 00:28:38]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3512kb
  • [2024-06-02 00:28:37]
  • 提交

answer

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

#define f first
#define s second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(v) v.begin(), v.end()
#define sz(v) (int)v.size()
#define sq(a) ((a)*(a))

#define MOO(i,a,b) for (int i=a; i<b; i++)
#define M00(i,a) for (int i=0; i<a; i++)
#define MOOd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define M00d(i,a) for (int i = (a)-1; i >= 0; i--)

#define per(i,a,b) for (int i = (b)-1; i >= a; i--)
#define rep(i,a,b) for (int i=a; i<b; i++)

#define FOR(i,a,b) for (int i=a; i<b; i++)
#define F0R(i,a) for (int i=0; i<a; i++)
#define ROF(i,a,b) for (int i = (b)-1; i >= a; i--)
#define R0F(i,a) for (int i = (a)-1; i >= 0; i--)

#define FAST ios::sync_with_stdio(0); cin.tie(0);
#define finish(x) return cout << x << endl, 0;
#define dbg(x) cerr << ">>> " << #x << " = " << x << endl;
#define _ << " _ " <<


template<class T> bool ckmin(T&a, T&b) { bool B = a > b; a = min(a,b); return B; }
template<class T> bool ckmax(T&a, T&b) { bool B = a < b; a = max(a,b); return B; }

typedef long double ld;
typedef pair<int,int> pi;
typedef pair<ld,ld> pld;
typedef complex<ld> cd;

typedef vector<int> vi;
typedef vector<ld> vld;
typedef vector<vector<int>> vvi;

const ld PI = acos(-1.0);
const ld EPS = 1e-7;
const int MOD = 1e9+7;

struct tnode {
    char c;
    int here;
    tnode* next[26];
	int cnt[26];
    tnode() {
        c = ' ';
        here = 0;
        M00(i, 26) next[i] = nullptr;
        M00(i, 26) cnt[i] = 0;
    }
};


void addToTrie(string s, tnode* root) {
    tnode* cur = root;
    for(char ch: s) {
        int idx = ch - 'a';
        if(cur->next[idx] == nullptr) {
            cur->next[idx] = new tnode();
        }
		cur->cnt[idx]++;
        cur = cur->next[idx];
        cur->c = ch;
    }
    cur->here += 1;
}

int unique(tnode* root) {
	int ans = root->here;
	M00(i, 26) ans += (root->cnt[i] > 0);
	return ans;
}

void free_trie(tnode* root) {
	if(root == nullptr) return;
	M00(i, 26) {
		free_trie(root->next[i]);
	}
	delete root;
}

void solve() {
	tnode* root = new tnode();
	int n, k; cin >> n >> k;
	vector<string> arr(n);
	M00(i, n) cin >> arr[i];
	M00(i, n) addToTrie(arr[i], root);

	tnode* rroot = root;
	vector<char> ans;
	while(true) {
		if(unique(root) >= k) {
			if(ans.size() == 0) {
				cout << "EMPTY";
			} else {
				for(auto &s : ans) cout << s;
			}
			cout << '\n';
			free_trie(rroot);
			return;
		}
		k -= root->here;
		int next = 0;
		int before = 0;
		M00(i, 26) next += (root->cnt[i] > 0);
		M00(i, 26) {
			next -= root->cnt[i] > 0;

			int poss = root->cnt[i] + next + before;
			if(poss >= k) {
				ans.pb('a' + i);
				k -= next;
				root = root->next[i];
				break;
			}
			before += root->cnt[i];
		}
	}
}

int32_t main() { FAST
	mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	int t; cin >> t;
	while(t--) solve();
}


詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3512kb

input:

2
5 3
gdcpc
gdcpcpcp
suasua
suas
sususua
3 3
a
b
c

output:

gdcpc
EMPTY

result:

ok 2 lines

Test #2:

score: -100
Time Limit Exceeded

input:

20000
5 3
apveofpr
irdbqk
rnionnjjrk
wrpihlsw
ofylfsriof
5 4
iiltghqg
kybhogptqf
jfnmqxzrdq
rpztcluq
tzmnrcjae
5 5
ysp
vujmkkcutl
ksawqeiiaf
waukilaq
wmlsutlued
5 3
pikybt
yiqmqvtjeq
tyrsoihf
vnvjrxpus
cuubpacubb
5 2
rihuvikhg
twrsuoervz
ukiekoohw
eysqgndpf
sxswpeqtaf
5 5
vxzhicbp
nvtdbgqal
jilppvpt...

output:


result: