제출 #265546

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
265546		tko919	수열 (APIO14_sequence)	C++17	0 / 100	38 ms	7436 KiB

sequence

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;

//template
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(v) (v).begin(),(v).end()
typedef long long int ll;
const int inf = 0x3fffffff; const ll INF = 0x1fffffffffffffff; const double eps=1e-12;
template<typename T>inline bool chmax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T>inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
//end

class CHT {
   struct F{
      ll a,b; int idx;
      F(ll a,ll b,int idx):a(a),b(b),idx(idx){}
   };
   deque<F> deq;
   bool check(F &f1,F &f2,F &f3) {
      return (f2.a - f1.a) * (f3.b - f2.b) >= (f2.b - f1.b) * (f3.a - f2.a);
   }
   ll f(F &f1,ll x) {return f1.a * x + f1.b;}
public:
   void add_line(ll a,ll b,int idx) {
      F f1 = F(a,b,idx);
      while(deq.size() >= 2 and check(deq[deq.size()-2], deq[deq.size()-1], f1)) {
         deq.pop_back();
      } deq.push_back(f1);
   }
   pair<ll,int> query(ll x) {
      while(deq.size() >= 2 and f(deq[0], x) <= f(deq[1], x)) {
         deq.pop_front();
      } return {f(deq[0], x),deq[0].idx};
   }
};

ll dp[210][101010]={},pre[210][101010];

int main(){
   int n,k; cin>>n>>k;
   vector<ll> a(n),rui(n+1);
   rep(i,0,n){
      cin>>a[i]; rui[i+1]=a[i];
   }
   rep(i,0,n)rui[i+1]+=rui[i];
   rep(i,0,k+1)rep(j,0,n+1)dp[i][j]=-INF;
   dp[0][0]=0;
   /*
   rep(i,1,k+1)rep(j,1,n)rep(i2,0,j)if(dp[i-1][i2]>=0){
      ll add=(rui[j]-rui[i2])*(rui[n]-rui[j]);
      if(chmax(dp[i][j],dp[i-1][i2]+add))pre[i][j]=i2;
   }
   */
   rep(i,1,k+1){
      CHT cht;
      rep(j,1,n){
         if(dp[i-1][j-1]>=0)cht.add_line(-rui[j-1],dp[i-1][j-1],j-1);
         auto [v,x]=cht.query(rui[n]-rui[j]);
         if(chmax(dp[i][j],v+rui[j]*(rui[n]-rui[j])))pre[i][j]=x;
      }
   }
   ll res=-INF; int pos=-1;
   rep(i,0,n)if(chmax(res,dp[k][i]))pos=i;
   cout<<res<<endl;
   int cur=k; vector<int> ps;
   while(cur){
      ps.push_back(pos);
      pos=pre[cur][pos]; cur--;
   }
   reverse(ALL(ps));
   rep(i,0,k)cout<<ps[i]<<endl;
   return 0;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29