이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <complex>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <unordered_set>
#include <vector>
using namespace std;
// BEGIN NO SAD
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define derr if(1) cerr
typedef vector<int> vi;
// END NO SAD
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<vector<ll>> matrix;
typedef pair<int, pii> state;
ll parse() {
string s;
cin >> s;
ll ret = 0;
for(int i = 0; i < sz(s); i++) {
ret <<= 1;
ret |= s[i] == '1';
}
return ret;
}
vector<ll> bad;
int n, k;
bool finite(ll src, ll snk) {
set<ll> s;
vector<ll> all;
all.reserve(n*k+n+1);
all.push_back(src);
s.insert(src);
for(int i = 0; i < sz(all) && sz(all) <= n*k; i++) {
for(int j = 0; j < n; j++) {
ll curr = all[i] ^ (1LL << j);
if(s.count(curr)) continue;
auto it = lower_bound(all(bad), curr);
if(it != bad.end() && *it == curr) continue;
s.insert(curr);
all.push_back(curr);
}
}
return !s.count(snk) && sz(all) <= n*k;
}
void solve() {
cin >> n >> k;
ll src = parse();
ll snk = parse();
bad.reserve(k);
for(int i = 0; i < k; i++) {
bad.push_back(parse());
}
sort(all(bad));
if(finite(src, snk) || finite(snk, src)) cout << "NIE\n";
else cout << "TAK\n";
}
// are there edge cases (N=1?)
// are array sizes proper (scaled by proper constant, for example 2* for koosaga tree)
// integer overflow?
// DS reset properly between test cases
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL); cout.tie(NULL);
solve();
}
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