結果
問題 | No.1029 JJOOII 3 |
ユーザー | hirono999 |
提出日時 | 2020-04-18 21:20:30 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 319 ms / 2,000 ms |
コード長 | 5,069 bytes |
コンパイル時間 | 2,143 ms |
コンパイル使用メモリ | 224,552 KB |
実行使用メモリ | 8,960 KB |
最終ジャッジ日時 | 2024-10-04 00:21:17 |
合計ジャッジ時間 | 10,668 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 16 ms
8,704 KB |
testcase_01 | AC | 10 ms
8,960 KB |
testcase_02 | AC | 22 ms
8,704 KB |
testcase_03 | AC | 64 ms
8,576 KB |
testcase_04 | AC | 64 ms
8,704 KB |
testcase_05 | AC | 242 ms
8,832 KB |
testcase_06 | AC | 247 ms
8,704 KB |
testcase_07 | AC | 239 ms
8,832 KB |
testcase_08 | AC | 242 ms
8,796 KB |
testcase_09 | AC | 238 ms
8,832 KB |
testcase_10 | AC | 232 ms
8,832 KB |
testcase_11 | AC | 241 ms
8,704 KB |
testcase_12 | AC | 246 ms
8,832 KB |
testcase_13 | AC | 237 ms
8,832 KB |
testcase_14 | AC | 246 ms
8,832 KB |
testcase_15 | AC | 212 ms
8,960 KB |
testcase_16 | AC | 219 ms
8,960 KB |
testcase_17 | AC | 204 ms
8,960 KB |
testcase_18 | AC | 231 ms
8,820 KB |
testcase_19 | AC | 22 ms
8,704 KB |
testcase_20 | AC | 164 ms
8,832 KB |
testcase_21 | AC | 161 ms
8,832 KB |
testcase_22 | AC | 161 ms
8,728 KB |
testcase_23 | AC | 163 ms
8,832 KB |
testcase_24 | AC | 162 ms
8,832 KB |
testcase_25 | AC | 168 ms
8,704 KB |
testcase_26 | AC | 228 ms
8,800 KB |
testcase_27 | AC | 257 ms
8,832 KB |
testcase_28 | AC | 237 ms
8,812 KB |
testcase_29 | AC | 233 ms
8,832 KB |
testcase_30 | AC | 247 ms
8,832 KB |
testcase_31 | AC | 236 ms
8,704 KB |
testcase_32 | AC | 314 ms
8,704 KB |
testcase_33 | AC | 315 ms
8,624 KB |
testcase_34 | AC | 319 ms
8,704 KB |
testcase_35 | AC | 317 ms
8,704 KB |
testcase_36 | AC | 315 ms
8,704 KB |
testcase_37 | AC | 11 ms
8,704 KB |
testcase_38 | AC | 16 ms
8,576 KB |
testcase_39 | AC | 19 ms
8,704 KB |
testcase_40 | AC | 19 ms
8,704 KB |
ソースコード
#include <bits/stdc++.h> const long long INF = 1LL << 60; const long long MOD = 1000000007; #define rep(i, n) for (ll i = 0; i < (n); ++i) #define rep1(i, n) for (ll i = 1; i <= (n); ++i) #define rrep(i, n) for (ll i = (n - 1); i >= 0; --i) #define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c))) #define ALL(obj) (obj).begin(), (obj).end() #define RALL(obj) (obj).rbegin(), (obj).rend() #define pb push_back #define to_s to_string #define len(v) (ll)v.size() #define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) #define print(x) cout << (x) << '\n' #define drop(x) cout << (x) << '\n', exit(0) #define debug(x) cout << #x << ": " << (x) << '\n' using namespace std; using ll = long long; typedef pair<ll, ll> P; typedef vector<ll> vec; typedef vector<vector<ll>> vec2; typedef vector<vector<vector<ll>>> vec3; template<class S, class T> inline bool chmax(S &a, const T &b) { if (a<b) { a=b; return 1; } return 0; } template<class S, class T> inline bool chmin(S &a, const T &b) { if (b<a) { a=b; return 1; } return 0; } inline ll msb(ll v) { return 1 << (31 - __builtin_clzll(v)); } inline ll devc(ll x, ll y) { return (x + y - 1) / y; } struct IoSetup { IoSetup() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(10); cerr << fixed << setprecision(10); } } iosetup; template< typename T1, typename T2 > ostream &operator << (ostream &os, const pair< T1, T2 > &p) { os << p.first << " " << p.second; return os; } template< typename T1, typename T2 > istream &operator >> (istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; } template< typename T > ostream &operator << (ostream &os, const vector< T > &v){ for (int i = 0; i < (int)v.size(); ++i) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } template< typename T > istream &operator >> (istream &is, vector< T > &v){ for(T &in : v) is >> in; return is; } /*--------------------------------- Tools ------------------------------------------*/ template< typename T > vector<T> cumsum(const vector<T> &X){ vector<T> res(X.size() + 1, 0); for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i]; return res; } //f : 単調性の関数 template< typename S, typename T, typename F> pair<T, T> bisearch(S left, T right, F f) { while(abs(right - left) > 1){ T mid = (right + left) / 2; if(f(mid)) right = mid; else left = mid; } return {left, right}; } template< typename S, typename T, typename F> double trisearch(S left, T right, F f, double precision = 1e-10){ double low = left, high = right; while(high - low > precision){ double mid_left = high / 3 + low * 2 / 3; double mid_right = high * 2 / 3 + low / 3; if(f(mid_left) >= f(mid_right)) low = mid_left; else high = mid_right; } return high; } /*------------------------------- Main Code Here -----------------------------------------*/ int main() { ll N, K; cin >> N >> K; vector<pair<string, ll>> S(N); cin >> S; vec2 f(1e5 + 1, vec(3, INF)); //0 : J, 1 : O, 2 : I f[0][0] = f[0][1] = f[0][2] = 0; rep(i, N){ map<char, ll> mp; auto [s, v] = S[i]; for(auto c : s) ++mp[c]; for (ll j = 0; j <= 1e5; ++j){ if(j >= mp['J']) chmin(f[j][0], f[j - mp['J']][0] + v); if(j >= mp['O']) chmin(f[j][1], f[j - mp['O']][1] + v); if(j >= mp['I']) chmin(f[j][2], f[j - mp['I']][2] + v); } for (ll j = 1e5; j >= 1; --j) rep(k, 3) chmin(f[j - 1][k], f[j][k]); } if(f[1][0] == INF or f[1][1] == INF or f[1][2] == INF) drop(-1); ll ans = f[K][0] + f[K][1] + f[K][2]; rep(i, N) rep(j, N){ auto [s1, v1] = S[i]; auto [s2, v2] = S[j]; ll l1 = len(s1), l2 = len(s2); vec J1(l1), O1(l1), I1(l1); vec J2(l2), O2(l2), I2(l2); rep(k, l1){ J1[k] = s1[k] == 'J'; O1[k] = s1[k] == 'O'; I1[k] = s1[k] == 'I'; } rep(k, l2){ J2[k] = s2[k] == 'J'; O2[k] = s2[k] == 'O'; I2[k] = s2[k] == 'I'; } vec j1, o1, i1, j2, o2, i2; j1 = cumsum(J1); o1 = cumsum(O1); i1 = cumsum(I1); j2 = cumsum(J2); o2 = cumsum(O2); i2 = cumsum(I2); rep(x, l1 + 1) rep(y, l2 + 1){ ll ctj = j1[x]; ll cto = o1.back() - o1[x] + o2[y]; ll cti = i2.back() - i2[y]; chmin(ans, v1 + v2 + f[max(K - ctj, 0LL)][0] + f[max(K - cto, 0LL)][1] + f[max(K - cti, 0LL)][2]); } if(j == 0){ rep(x, l1 + 1) rep(y, l1 + 1) if(x < y) { ll ctj = j1[x]; ll cto = o1[y] - o1[x]; ll cti = i1.back() - i1[y]; if(cto >= K) chmin(ans, v1 + f[max(K - ctj, 0LL)][0] + f[max(K - cti, 0LL)][2]); } } } print(ans); return 0; }