結果
問題 | No.783 門松計画 |
ユーザー | はむこ |
提出日時 | 2017-07-18 17:37:39 |
言語 | C++11 (gcc 11.4.0) |
結果 |
TLE
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 7,785 bytes |
コンパイル時間 | 1,779 ms |
コンパイル使用メモリ | 175,896 KB |
実行使用メモリ | 8,824 KB |
最終ジャッジ日時 | 2024-07-23 12:16:45 |
合計ジャッジ時間 | 12,371 ms |
ジャッジサーバーID (参考情報) |
judge4 / 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 | 61 ms
8,712 KB |
testcase_01 | AC | 18 ms
8,680 KB |
testcase_02 | AC | 61 ms
8,784 KB |
testcase_03 | AC | 19 ms
8,648 KB |
testcase_04 | AC | 47 ms
8,520 KB |
testcase_05 | AC | 46 ms
8,688 KB |
testcase_06 | AC | 47 ms
8,632 KB |
testcase_07 | AC | 61 ms
8,680 KB |
testcase_08 | AC | 46 ms
8,644 KB |
testcase_09 | AC | 60 ms
8,664 KB |
testcase_10 | AC | 907 ms
8,684 KB |
testcase_11 | TLE | - |
testcase_12 | AC | 1,356 ms
8,780 KB |
testcase_13 | AC | 255 ms
8,668 KB |
testcase_14 | AC | 541 ms
8,760 KB |
testcase_15 | AC | 623 ms
8,752 KB |
testcase_16 | AC | 282 ms
8,652 KB |
testcase_17 | AC | 260 ms
8,672 KB |
testcase_18 | AC | 115 ms
8,684 KB |
testcase_19 | AC | 577 ms
8,684 KB |
testcase_20 | AC | 46 ms
8,688 KB |
testcase_21 | AC | 255 ms
8,732 KB |
testcase_22 | AC | 379 ms
8,784 KB |
testcase_23 | AC | 18 ms
8,644 KB |
testcase_24 | AC | 644 ms
8,760 KB |
testcase_25 | AC | 398 ms
8,784 KB |
testcase_26 | AC | 357 ms
8,736 KB |
ソースコード
#include <bits/stdc++.h> #include <sys/time.h> using namespace std; #define rep(i,n) for(long long i = 0; i < (long long)(n); i++) #define repi(i,a,b) for(long long i = (long long)(a); i < (long long)(b); i++) #define pb push_back #define all(x) (x).begin(), (x).end() #define fi first #define se second #define mt make_tuple #define mp make_pair #define ZERO(a) memset(a,0,sizeof(a)) template<class T1, class T2> bool chmin(T1 &a, T2 b) { return b < a && (a = b, true); } template<class T1, class T2> bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); } #define exists find_if #define forall all_of using ll = long long; using vll = vector<ll>; using vvll = vector<vll>; using P = pair<ll, ll>; using ld = long double; using vld = vector<ld>; using vi = vector<int>; using vvi = vector<vi>; vll conv(vi& v) { vll r(v.size()); rep(i, v.size()) r[i] = v[i]; return r; } using Pos = complex<double>; template <typename T, typename U> ostream &operator<<(ostream &o, const pair<T, U> &v) { o << "(" << v.first << ", " << v.second << ")"; return o; } template<size_t...> struct seq{}; template<size_t N, size_t... Is> struct gen_seq : gen_seq<N-1, N-1, Is...>{}; template<size_t... Is> struct gen_seq<0, Is...> : seq<Is...>{}; template<class Ch, class Tr, class Tuple, size_t... Is> void print_tuple(basic_ostream<Ch,Tr>& os, Tuple const& t, seq<Is...>){ using s = int[]; (void)s{0, (void(os << (Is == 0? "" : ", ") << get<Is>(t)), 0)...}; } template<class Ch, class Tr, class... Args> auto operator<<(basic_ostream<Ch, Tr>& os, tuple<Args...> const& t) -> basic_ostream<Ch, Tr>& { os << "("; print_tuple(os, t, gen_seq<sizeof...(Args)>()); return os << ")"; } ostream &operator<<(ostream &o, const vvll &v) { rep(i, v.size()) { rep(j, v[i].size()) o << v[i][j] << " "; o << endl; } return o; } template <typename T> ostream &operator<<(ostream &o, const vector<T> &v) { o << '['; rep(i, v.size()) o << v[i] << (i != v.size()-1 ? ", " : ""); o << "]"; return o; } template <typename T> ostream &operator<<(ostream &o, const set<T> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]"; return o; } template <typename T, typename U> ostream &operator<<(ostream &o, const map<T, U> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it << (next(it) != m.end() ? ", " : ""); o << "]"; return o; } template <typename T, typename U, typename V> ostream &operator<<(ostream &o, const unordered_map<T, U, V> &m) { o << '['; for (auto it = m.begin(); it != m.end(); it++) o << *it; o << "]"; return o; } vector<int> range(const int x, const int y) { vector<int> v(y - x + 1); iota(v.begin(), v.end(), x); return v; } template <typename T> istream& operator>>(istream& i, vector<T>& o) { rep(j, o.size()) i >> o[j]; return i;} string bits_to_string(ll input, ll n=64) { string s; rep(i, n) s += '0' + !!(input & (1ll << i)); reverse(all(s)); return s; } template <typename T> unordered_map<T, ll> counter(vector<T> vec){unordered_map<T, ll> ret; for (auto&& x : vec) ret[x]++; return ret;}; string substr(string s, P x) {return s.substr(x.fi, x.se - x.fi); } struct ci : public iterator<forward_iterator_tag, ll> { ll n; ci(const ll n) : n(n) { } bool operator==(const ci& x) { return n == x.n; } bool operator!=(const ci& x) { return !(*this == x); } ci &operator++() { n++; return *this; } ll operator*() const { return n; } }; size_t random_seed; namespace std { using argument_type = P; template<> struct hash<argument_type> { size_t operator()(argument_type const& x) const { size_t seed = random_seed; seed ^= hash<ll>{}(x.fi); seed ^= (hash<ll>{}(x.se) << 1); return seed; } }; }; // hash for various class namespace myhash{ const int Bsizes[]={3,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81}; const int xor_nums[]={0x100007d1,0x5ff049c9,0x14560859,0x07087fef,0x3e277d49,0x4dba1f17,0x709c5988,0x05904258,0x1aa71872,0x238819b3,0x7b002bb7,0x1cf91302,0x0012290a,0x1083576b,0x76473e49,0x3d86295b,0x20536814,0x08634f4d,0x115405e8,0x0e6359f2}; const int hash_key=xor_nums[rand()%20]; const int mod_key=xor_nums[rand()%20]; template <typename T> struct myhash{ std::size_t operator()(const T& val) const { return (hash<T>{}(val)%mod_key)^hash_key; } }; }; template <typename T> class uset:public std::unordered_set<T,myhash::myhash<T>> { using SET=std::unordered_set<T,myhash::myhash<T>>; public: uset():SET(){SET::rehash(myhash::Bsizes[rand()%20]);} }; template <typename T,typename U> class umap:public std::unordered_map<T,U,myhash::myhash<T>> { public: using MAP=std::unordered_map<T,U,myhash::myhash<T>>; umap():MAP(){MAP::rehash(myhash::Bsizes[rand()%20]);} }; struct timeval start; double sec() { struct timeval tv; gettimeofday(&tv, NULL); return (tv.tv_sec - start.tv_sec) + (tv.tv_usec - start.tv_usec) * 1e-6; } struct init_{init_(){ gettimeofday(&start, NULL); ios::sync_with_stdio(false); cin.tie(0); srand((unsigned int)time(NULL)); random_seed = RAND_MAX / 2 + rand() / 2; }} init__; static const double EPS = 1e-14; static const long long INF = 1e18; static const long long mo = 1e9+7; #define ldout fixed << setprecision(40) // 複数個買えるのに注意 // lを長さ2500の数列に展開する必要? // 相違なら必ず構成できることがわかる。 // // 相違でない場合は? // // DP: i個見て、j円使っていて、門松列々になる? // 被っている場合は、一番安いのだけを残せば良い。 // // 最も重複しているものが、n/2個以下ならばよい? // うーん、2500に展開するのが楽そうだなあ... // DP[i][pre][ppre][b][k] = ちょうどi個見て、直前が長さpreで、直前前が長さppre、直前前々が存在して(b)、k円使っている場合の最大長さ // 遷移は、次を取るか取らないか // O(n^3 c) ll dp[2][55][55][2][55] = {}; int main(void) { ll n, c; cin >> n >> c; assert(1 <= n && n <= 50); assert(1 <= c && c <= 50); map<ll, ll> memo; vll l_(n), w_(n); cin >> l_ >> w_; rep(i, n) { assert(1 <= l_[i] && l_[i] <= 50); assert(1 <= w_[i] && w_[i] <= 50); } rep(i, n) { ll u = l_[i], v = w_[i]; if (memo.count(u)) { chmax(memo[u], v); } else { memo[u] = v; } } vll l, w; for (auto x : memo) { l.pb(x.fi); w.pb(x.se); } n = l.size(); rep(_, 50) rep(i, n) { l.push_back(l[i]); w.push_back(w[i]); } n = l.size(); rep(i, 2) rep(j, 55) rep(h, 55) rep(b, 2) rep(k, 55) dp[i][j][h][b][k] = -INF; const int init = 54; dp[0][init][init][0][0] = 0; rep(i, n) { ll curr = i % 2; ll next = (i + 1) % 2; rep(pre, 55) rep(ppre, 55) rep(b, 2) rep(k, 55) { if (dp[curr][pre][ppre][b][k] == -INF) continue; // 配る必要なし chmax(dp[next][pre][ppre][b][k], dp[curr][pre][ppre][b][k]); // 買わない if (k+w[i]>c) continue; // コストオーバー if ( (ppre == init && pre == init) || (ppre == init && pre != init && pre != l[i]) || (ppre != init && pre != init && (ppre < pre && pre > l[i] && l[i] != ppre)) || (ppre != init && pre != init && (ppre > pre && pre < l[i] && l[i] != ppre)) ) { // 門松 chmax(dp[next][l[i]][pre][ppre!=init][k+w[i]], dp[curr][pre][ppre][b][k] + l[i]); // 買えるし門松なので買う } } } ll ret = -INF; rep(pre, 55) rep(ppre, 55) rep(k, 55) { chmax(ret, dp[n%2][pre][ppre][1][k]); } if (ret == -INF) { cout << 0 << endl; } else { cout << ret << endl; } return 0; }