結果

問題 No.1029 JJOOII 3
ユーザー emthrm
提出日時 2020-04-17 22:05:52
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 2,032 bytes
コンパイル時間 12,766 ms
コンパイル使用メモリ 289,452 KB
最終ジャッジ日時 2025-01-09 20:05:53
ジャッジサーバーID
(参考情報)
judge3 / judge2
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 33 TLE * 5
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function 'int main()':
main.cpp:63:16: warning: 'status' may be used uninitialized [-Wmaybe-uninitialized]
   63 |         status = status + 1;
      |         ~~~~~~~^~~~~~~~~~~~
main.cpp:50:9: note: 'status' was declared here
   50 |     int status, cnt = i % k;
      |         ^~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
IOSetup() {
cin.tie(nullptr);
ios_base::sync_with_stdio(false);
cout << fixed << setprecision(20);
}
} iosetup;
int main() {
int n, k; cin >> n >> k;
vector<string> s(n);
vector<int> c(n);
REP(i, n) cin >> s[i] >> c[i];
{
bool j = false, o = false, I = false;
REP(i, n) for (char c : s[i]) {
j |= c == 'J';
o |= c == 'O';
I |= c == 'I';
}
// cout << j << ' ' << o << ' ' << i << '\n';
if (!j || !o || !I) {
cout << "-1\n";
return 0;
}
}
vector<ll> dp(k * 3 + 1, LINF);
dp[0] = 0;
REP(i, k * 3) REP(j, n) {
int status, cnt = i % k;
if (i < k) status = 0;
if (k <= i && i < k * 2) status = 1;
if (k * 2 <= i) status = 2;
for (char c : s[j]) {
if (status == 0) {
if (c == 'J') ++cnt;
} else if (status == 1) {
if (c == 'O') ++cnt;
} else if (status == 2) {
if (c == 'I') ++cnt;
}
if (cnt == k && status < 2) {
status = status + 1;
cnt = 0;
}
}
chmin(cnt, k);
chmin(dp[status * k + cnt], dp[i] + c[j]);
}
cout << dp[k * 3] << '\n';
return 0;
}
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