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Cod sursa(job #772420)
#include <iostream>
#include <fstream>
#include <deque>
using namespace std;
ifstream f("ferma.in");
ofstream g("ferma.out");
#define nmax 10005
#define Kmax 1005
int n, K, a[2*nmax], dp[Kmax][2*nmax], s[2*nmax];
deque<int> dq;
void citeste(){
f >> n >> K;
for(int i=1; i<=n; i++) f >> a[i], s[i] = s[i-1]+a[i];
for(int i=n+1; i<=2*n; i++) a[i] = a[i-n], s[i] = s[i-1]+a[i];//dublez vectorul pentru a fi circular
// for(int i=1; i<=2*n; i++) cout << s[i] << " ";
// cout << "\n";
}
int afla_poz(){
int poz = 0;
int rez = -2000000000;
dq.clear();
int dr = 0;
for(int i=1; i<=2*n; i++){
while(dq.size() && i-dq.front()+1 > n) dq.pop_front();//scot secventele mai mari ca si n;
while(dq.size() && s[i] <= s[dq.back()]) dq.pop_back();//am nevoie de cea mai mica secventa
dq.push_back(i);
if (rez < s[i] - s[dq.front()]){
rez = s[i] - s[dq.front()];
poz = dq.front() + 1;
dr = i;
}
}
return poz;
}
void rezolva(){
//dp[i][j] = productivitate maxima din primele j gaini avand i treceri
int rez = 0;
int poz = afla_poz();
for(int i=0; i<=K; i++){
for(int j=1; j<=n; j++) dp[i][j] = 0;
}
dp[1][poz] = max(0, a[poz]);
for(int i=1; i<=K; i++){//am i treceri
for(int j=poz+1; j<=poz+n-1; j++){
dp[i][j] = dp[i][j-1];
for(int k=j; k>=poz; k--){//incerc fiecare secventa de genul [j,k]; [j,k-1]; .. [j,poz]
dp[i][j] = max(dp[i][j], (s[j] - s[k-1]) + dp[i-1][k-1]);
}
}
}
rez = dp[K][poz+n-1];
g << rez << "\n";
}
int main(){
citeste();
rezolva();
f.close();
g.close();
return 0;
}