結果
| 問題 |
No.3097 Azuki Kurai
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2025-03-29 04:50:10 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 2,449 bytes |
| コンパイル時間 | 3,740 ms |
| コンパイル使用メモリ | 222,460 KB |
| 実行使用メモリ | 20,736 KB |
| 最終ジャッジ日時 | 2025-03-29 04:50:25 |
| 合計ジャッジ時間 | 11,982 ms |
|
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 |
|---|---|
| other | AC * 2 WA * 30 |
ソースコード
#include <bits/stdc++.h>
#define rep(i, p, n) for (ll i = p; i < (ll)(n); i++)
#define rep2(i, p, n) for (ll i = p; i >= (ll)(n); i--)
using namespace std;
using ll = long long;
using ld = long double;
const double pi = 3.141592653589793;
const long long inf = 2 * 1e9;
const long long linf = 4 * 1e18;
const ll mod1 = 1000000007;
const ll mod2 = 998244353;
// atcoder
#include <atcoder/all>
using namespace atcoder;
using mint1 = modint1000000007;
using mint2 = modint998244353;
constexpr int MAX_N = 10;
constexpr int MAX_M = 2000;
constexpr int MAX_STATES = (1 << MAX_N);
array<pair<ll, ll>, 4> base = {{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}};
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
ll N, M, K;
cin >> N >> M >> K;
ll A[MAX_N];
rep(i, 0, N) {
cin >> A[i];
}
ll dp[MAX_M + 1][MAX_STATES];
rep(i, 0, M + 1) rep(j, 0, MAX_STATES) dp[i][j] = linf;
rep(i, 0, (1 << N)) {
dp[0][i] = 0;
rep(j, 0, N) {
if (!(i & (1 << j))) {
dp[0][i] += A[j];
}
}
}
ll pl[MAX_STATES][MAX_N + 1];
ll mae[MAX_N][MAX_STATES][MAX_N + 1];
rep(i, 0, (1 << N)) {
int F = 0;
for (ll c = i; c; c = (c - 1) & i) {
pl[i][F++] = (__builtin_popcount(i) - __builtin_popcount(c)) * K;
}
pl[i][F] = (__builtin_popcount(i) - __builtin_popcount(0)) * K;
}
rep(p, 0, N) {
rep(i, 0, (1 << N)) {
int F = 0;
for (ll c = i; c; c = (c - 1) & i) {
ll now = i;
now |= ((c * 2) - ((c * 2) & (1 << N)));
now |= (((c * 2) & (1 << N)) >> N);
now |= (c / 2);
now |= (c % 2) * (1 << (N - 1));
now -= (now & (1 << p));
mae[p][i][F++] = now;
}
ll now = i;
now -= (now & (1 << p));
mae[p][i][F] = now;
}
}
rep(p, 0, M) {
ll B;
cin >> B;
B--;
rep(i, 0, (1 << N)) {
ll F = 0;
for (ll c = i; c; c = (c - 1) & i) {
dp[p + 1][mae[B][i][F]] = min(dp[p + 1][mae[B][i][F]], dp[p][i] + pl[i][F]);
F++;
}
dp[p + 1][mae[B][i][F]] = min(dp[p + 1][mae[B][i][F]], dp[p][i] + pl[i][F]);
}
}
rep(i, 0, M) {
cout << dp[i + 1][0] << endl;
}
}