結果
| 問題 |
No.1652 XOR Inequalities
|
| コンテスト | |
| ユーザー |
Kude
|
| 提出日時 | 2021-08-20 23:14:42 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 2,227 bytes |
| コンパイル時間 | 4,455 ms |
| コンパイル使用メモリ | 260,332 KB |
| 最終ジャッジ日時 | 2025-01-24 00:32:13 |
|
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 WA * 1 |
| other | AC * 10 WA * 43 |
ソースコード
#include<bits/stdc++.h>
#include<atcoder/all>
using namespace std;
using namespace atcoder;
#define rep(i,n)for (int i = 0; i < int(n); ++i)
#define rrep(i,n)for (int i = int(n)-1; i >= 0; --i)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
template<class T> void chmax(T& a, const T& b) {a = max(a, b);}
template<class T> void chmin(T& a, const T& b) {a = min(a, b);}
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
const unsigned long long INF = 1002003004005006007;
template<class T>
std::vector<unsigned long long> dijkstra(std::vector<std::vector<std::pair<int,T>>>& to, int s=0) {
struct QueElem {
int v;
unsigned long long c;
bool operator<(const QueElem a) const {return c > a.c;}
QueElem(int v, unsigned long long c): v(v), c(c) {}
};
std::priority_queue<QueElem> q;
std::vector<unsigned long long> dist(to.size(), INF);
dist[s] = 0;
q.emplace(s, 0);
while(!q.empty()) {
QueElem qe = q.top(); q.pop();
int u = qe.v;
unsigned long long c = qe.c;
if (c > dist[u]) continue;
for(auto vc: to[u]) {
int v = vc.first;
unsigned long long nc = c + vc.second;
if (nc < dist[v]) {
dist[v] = nc;
q.emplace(v, nc);
}
}
}
return dist;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
int n2 = 1 << n;
VI a(n2);
rep(i, n2) cin >> a[i];
vector<vector<P>> to(n2);
rep(i, n2) rep(j, n2) if (a[j] != -1) {
int k = i ^ j;
to[i].emplace_back(k, a[j]);
}
auto dist1 = dijkstra(to);
rep(i, n2) if (a[i] == -1) {
a[i] = (dist1[i] == INF ? (1 << 30) - 1 : dist1[i]);
}
to = vector<vector<P>>(n2);
rep(i, n2) rep(j, n2) if (a[j] != -1) {
int k = i ^ j;
to[i].emplace_back(k, a[j]);
}
auto dist2 = dijkstra(to);
rep(i, n2) if (i && dist2[i] < a[i]) {
cout << "No" << endl;
return 0;
}
cout << "Yes" << endl;
rep(i, n2) cout << a[i] << " \n"[i + 1 == n2];
}
Kude