이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <math.h>
using namespace std;
#define pb push_back
#define INF INT32_MAX
#define vt vector
typedef vt<int> vi;
typedef pair<int,int> pii;
typedef tuple<int,int,int> tiii;
typedef long long ll;
#define MOD 1000000007
ll delivery(int n, int k, int l, int arr[]){
if(k==1){
ll ans = 0;
for(int i = 0; i < n; i++){
ans+=2*min(arr[i],l-arr[i]);
}
return ans;
}else if(k==n){
if(arr[n-1]<=l/2){
return arr[n-1]*2;
}else if(arr[0]>l/2){
return 2* (l-arr[0]);
}else{
int Max1 = 0;
int Max2 = 0;
int i = 0;
for(; i < n; i++){
if(arr[i]<=l/2){
Max1 = max(Max1, arr[i]);
}else{
break;
}
}
Max2 = arr[i];
return min(l,2*(Max1+l-Max2));
}
}else{
vt<ll> vals1(n+1);
vt<ll> vals2(n+1);
vt<ll> dp(n+1);
dp[0] = 0;
multiset<ll> pq1;
pq1.insert(dp[0]-2*arr[0]);
vals1[0]=dp[0]-2*arr[0];
multiset<ll> pq2;
pq2.insert(0);
vals2[0] = 0;
for(int i = 1; i <= n; i++){
dp[i] = *(pq1.begin())+2*l;
dp[i] = min(dp[i],min(*(pq2.begin())+2*arr[i-1],*(pq2.begin())+l));
if(i!=n) {
vals1[i] = dp[i] - 2 * arr[i];
vals2[i] = dp[i];
pq1.insert(vals1[i]);
pq2.insert(vals2[i]);
}
if(i-k>=0){
pq1.erase(pq1.find(vals1[i-k]));
pq2.erase(pq2.find(vals2[i-k]));
}
}
return dp[n];
}
}
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