이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include <fstream>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("O3","unroll-loops")
using namespace __gnu_pbds;
using namespace std;
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
struct custom_hash {static uint64_t splitmix64(uint64_t x) {x += 0x9e3779b97f4a7c15;x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;x = (x ^ (x >> 27)) * 0x94d049bb133111eb;return x ^ (x >> 31);}size_t operator()(uint64_t x) const {static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();return splitmix64(x + FIXED_RANDOM);}};
// string __fname = ""; ifstream in (__fname + ".in"); ofstream out (__fname + ".out");
// #define cin in
// #define cout out
#define int64 long long
#define uint64 unsigned long long
#define x first
#define y second
#define pb push_back
#define pii pair <int, int>
#define pii64 pair <int64, int64>
#define db(x) cout << "> " << #x << ": " << (x) << "\n"
#define qr queries()
#define yn(x) if (x) {ctn("Yes");}else {ctn("No");}
void solve(int);
void queries(){int n;cin >> n;for (int i = 1; i <= n; i++) solve(i);}
template<class T>T ceildiv(T a, T b) {return a / b + !!(a % b);}
template<class T>T gcd (T a, T b){return (b ? gcd(b, a % b): a);}
template<class T>T lcm (T a, T b){return a * b / gcd(a, b);}
// // // // // // // // // // // // // // // // // // // // // //
/* TEMPLATE - VANILLA */
// // // // // // // // // // // // // // // // // // // // // //
const int ddx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int ddy[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
const double pi = 3.14159265359;
const double eps = 1e-6;
const int64 hash_inv = 940594066;
const int64 hash_p = 101;
const int64 mod = 1e9 + 7;
const int maxn = 20;
const int maxv = 1e3 + 2;
vector <int> dp[maxv];
bitset <1 << maxn> bts [maxn]; // bst[i][j] -> build first i elements with state j
int a[maxn];
int b[maxn];
int used;
int n,m;
void solve(int id){
return;
}
void back (int step) {
if (step == n) {
cout << "YES\n";
exit(0);
}
for (int v: dp[a[step]]) {
if (v & used) continue;
used+=v;
back(step + 1);
used-=v;
}
}
int main(){
ios_base::sync_with_stdio(0); cin.tie(0); cout << fixed; cout << setprecision(10);
cin >> n >> m;
for (int i = 0; i < n; i++) cin >> a[i];
for (int i = 0; i < m; i++) cin >> b[i];
dp[0].push_back(0);
for (int i = 0; i < m; i++){
for (int j = maxv - 1; j - b[i] >= 0; j--){
for (auto k: dp[j - b[i]]) dp[j].push_back(k + (1 << i));
}
}
for (int mask = 0; mask < (1 << maxn); mask++) {
for (int j: dp[0]) {
if (bts[0][mask]) continue;
if ((mask & j) == j) bts[0][mask] = 1;
}
}
for (int i = 0; i < n; i++){
for (int mask = 0; mask < (1 << maxn); mask++) {
for (int j: dp[i]) {
if (bts[i][mask]) continue;
if ((mask & j) == j && bts[i-1][mask - j]) bts[i][mask] = 1;
}
}
}
for (int i = 0; i < (1 << maxn); i++){
if (bts[n-1][i]) {
cout << "YES";
return 0;
}
}
// back(0);
cout << "NO";
return 0;
}
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