結果

問題 No.2026 Yet Another Knapsack Problem
ユーザー snuke
提出日時 2022-07-30 11:44:47
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 974 ms / 10,000 ms
コード長 3,623 bytes
コンパイル時間 2,395 ms
コンパイル使用メモリ 207,916 KB
最終ジャッジ日時 2025-01-30 16:34:27
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 42
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 getInt()’:
main.cpp:25:25: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   25 | int getInt(){int x;scanf("%d",&x);return x;}
      |                    ~~~~~^~~~~~~~~
main.cpp: In member function ‘void Solver::solve()’:
main.cpp:69:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   69 |     scanf("%d",&n);
      |     ~~~~~^~~~~~~~~

ソースコード

diff # 

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rep1(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < (t); ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define pob pop_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define newline puts("")
using namespace std;
template<typename T> using vc = vector<T>;
template<typename T> using vv = vc<vc<T>>;
template<typename T> using PQ = priority_queue<T,vc<T>,greater<T>>;
using uint = unsigned; using ull = unsigned long long;
using vi = vc<int>; using vvi = vv<int>; using vvvi = vv<vi>;
using ll = long long; using vl = vc<ll>; using vvl = vv<ll>; using vvvl = vv<vl>;
using P = pair<int,int>; using vp = vc<P>; using vvp = vv<P>;
int getInt(){int x;scanf("%d",&x);return x;}
vi pm(int n, int s=0) { vi a(n); iota(rng(a),s); return a;}
template<typename T>istream& operator>>(istream&i,vc<T>&v){rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const T&v,const string& d=""){stringstream s;rep(i,sz(v))(i?s<<d:s)<<v[i];return s.str();}
template<typename T>ostream& operator<<(ostream&o,const vc<T>&v){if(sz(v))o<<join(v," ");return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v){return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v){return o<<v.fi<<","<<v.se;}
template<typename T1,typename T2>bool mins(T1& x,const T2&y){if(x>y){x=y;return true;}else return false;}
template<typename T1,typename T2>bool maxs(T1& x,const T2&y){if(x<y){x=y;return true;}else return false;}
template<typename Tx, typename Ty>Tx dup(Tx x, Ty y){return (x+y-1)/y;}
template<typename T>ll suma(const vc<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
template<typename T>ll suma(const vv<T>&a){ll res(0);for(auto&&x:a)res+=suma(x);return res;}
template<typename T>void uni(T& a){sort(rng(a));a.erase(unique(rng(a)),a.end());}
template<typename T>void prepend(vc<T>&a,const T&x){a.insert(a.begin(),x);}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return;}
#define yes { puts("Yes"); return;}
#define no { puts("No"); return;}
#define ret(x) { cout<<(x)<<endl; return;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;

// sliding queue
template<typename T=ll>
struct SQ { // max
  using TP = pair<T,int>;
  deque<TP> q;
  void push(int i, T x) {
    while (sz(q) && q.back().fi <= x) q.pop_back();
    q.pb(TP(x,i));
  }
  T get(int i) { // i <= pos
    while (sz(q) && q[0].se < i) q.pop_front();
    if (sz(q)) return q[0].fi;
    return -LINF; // INF!
  }
};
//

struct Solver {
  void solve() {
    int n;
    scanf("%d",&n);
    vi c(n+1), a(n+1);
    rep1(i,n) cin>>c[i]>>a[i];
    vvl dp(1,vl(n+1));
    for (int x = n; x >= 1; --x) {
      ll na = a[x];
      int m = n/x;
      vvl p(m+1,vl(n+1,-LINF)); swap(dp,p);
      rep(sj,n+1) {
        int i = 0, j = sj;
        SQ q;
        while (i <= m && j <= n) {
          if (i < sz(p)) q.push(i,p[i][j]-na*i);
          ll now = q.get(i-c[x]);
          if (now != -LINF) dp[i][j] = now+na*i;
          i++; j += x;
        }
      }
    }
    rep1(k,n) {
      ll ans = dp[k][n];
      printf("%lld\n",ans);
    }
  }
};

int main() {
  int ts = 1;
  // scanf("%d",&ts);
  rep1(ti,ts) {
    Solver solver;
    solver.solve();
  }
  return 0;
}
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