제출 #270019

#제출 시각아이디문제언어결과실행 시간메모리
270019anubhavdhar수열 (APIO14_sequence)C++14
0 / 100
115 ms131076 KiB
#include<bits/stdc++.h> #define ll long long #define pb push_back #define mp make_pair #define pii pair<int, int> #define pll pair<ll, ll> #define ff first #define ss second #define vi vector<int> #define vl vector<ll> #define vii vector<pii> #define vll vector<pll> #define FOR(i,N) for(i=0;i<(N);++i) #define FORe(i,N) for(i=1;i<=(N);++i) #define FORr(i,a,b) for(i=(a);i<(b);++i) #define FORrev(i,N) for(i=(N);i>=0;--i) #define F0R(i,N) for(int i=0;i<(N);++i) #define F0Re(i,N) for(int i=1;i<=(N);++i) #define F0Rr(i,a,b) for(ll i=(a);i<(b);++i) #define F0Rrev(i,N) for(int i=(N);i>=0;--i) #define all(v) (v).begin(),(v).end() #define dbgLine cerr<<" LINE : "<<__LINE__<<"\n" #define ldd long double using namespace std; const int Alp = 26; const int __PRECISION = 9; const int inf = 1e9 + 8; const ldd PI = acos(-1); const ldd EPS = 1e-7; const ll MOD = 1e9 + 7; const ll MAXN = 1e5 + 5; const ll ROOTN = 320; const ll LOGN = 18; const ll INF = 1e18 + 1022; ll dp[MAXN][201], pre[MAXN], par[MAXN][201]; int N, K; struct line { mutable ll m, c, p, ind; bool operator < (const line & o) const {return m < o.m;} bool operator < (ll x) const {return p < x;} }; struct CHT : multiset<line, less<>> { ll div(ll a,ll b) { return a/b - ((a^b) < 0 && a%b);} bool isect(iterator x, iterator y) { if(y == end()) {x->p = INF; return false;} if(x->m == y-> m) x->p = (x->c > y->c ) ? INF: -INF; else x->p = div((y->c - x->c), (x->m - y->m)); return (x->p >= y->p); } void add(ll sl, ll in, ll i) { auto z = insert({sl, in, 0, i}), y = z++, x = y; while(isect(y, z)) z = erase(z); if(x != begin() && isect(--x,y)) isect(x, y = erase(y)); while((y=x) != begin() && (--x)->p >= y->p) isect(x,erase(y)); } pll query(ll x) { assert(!empty()); auto l = *lower_bound(x); return mp(l.m * x + l.c, l.ind); } }box[201]; signed main() { /* ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); */ ll p, q = INF; cin>>N>>K; pre[0] = 0; F0Re(i, N) cin>>p, pre[i] = p + pre[i-1]; p = 0; F0Re(i, N) cout<<pre[i]<<' '; cout<<endl; F0R(i, N+1) F0R(j, K+1) dp[i][j] = par[i][j] = 0; // F0Rr(k, j-1, i) dp[i][j] = max(dp[k][j-1] + pre[N]*pre[i] - pre[i]*pre[i] - pre[N]*pre[k] + pre[i]*pre[k]), dp[i][j]); F0Re(i, N) { F0Re(j, min(i, K+1)) { if(j == 1) dp[i][j] = pre[i]*(pre[N] - pre[i]); else {pll v = box[j-1].query(pre[i]); dp[i][j] = pre[N]*pre[i] - pre[i]*pre[i] + v.ff; par[i][j] = v.ss;} // p = max(p, dp[i][j]); } p = max(p, dp[i][K]); F0Re(j,min(i, K+1)) box[j].add(pre[i], dp[i][j] - pre[i] * pre[N], i); } F0Re(i, N) F0Re(j, K) p = max(p, dp[i][j]);//, cout<<"dp["<<i<<"]["<<j<<"] = "<<dp[i][j]<<", par["<<i<<"]["<<j<<"] = "<<par[i][j]<<'\n'; cout<<p<<'\n'; F0Re(i, N) if(dp[i][K] == p) { q = i; break; } while(q and K) { cout<<q<<' '; q = par[q][K--]; } return 0; }
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