Submission #470721

#	Time	Username	Problem	Language	Result	Execution time	Memory
470721		Newtech66	K blocks (IZhO14_blocks)	C++17	0 / 100	1 ms	312 KiB

blocks

#include<bits/stdc++.h>
using namespace std;
using lol=long long int;
#define endl "\n"
const lol mod1=1e9+7,mod2=998244353;
const lol inf=1e18+8;
const double eps=1e-12;
const int N=1e5+5;
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int _=1;
//cin>>_;
while(_--)
{
    int n,k;
    cin>>n>>k;
    vector<lol> v(n),pmax(n);
    for(auto& e:v)  cin>>e;
    pmax[0]=v[0];
    for(int i=1;i<n;i++)    pmax[i]=max(pmax[i-1],v[i]);
    //int dp[n][k][2] //pos, blocks, take last or start new
    vector<vector<vector<pair<lol,lol>>>> dp(n,vector<vector<pair<lol,lol>>>(k,vector<pair<lol,lol>>(2)));
    dp[0][0][0]={v[0],v[0]};
    dp[0][0][1]={inf,inf};
    for(int i=1;i<k;i++)
    {
        dp[0][i][0]=dp[0][i][1]={inf,inf};
    }
    for(int i=1;i<n;i++)
    {
        dp[i][0][0]={pmax[i],pmax[i]};
        dp[i][0][1]={inf,inf};
        for(int j=1;j<k;j++)
        {
            lol a=dp[i-1][j][0].first+max(0ll,v[i]-dp[i-1][j][0].second),b=dp[i-1][j][1].first+max(0ll,v[i]-dp[i-1][j][1].second);
            if(a<b) dp[i][j][0]={a,max(v[i],dp[i-1][j][0].second)};
            else    dp[i][j][0]={b,max(v[i],dp[i-1][j][1].second)};
            dp[i][j][1]={min(dp[i-1][j-1][0].first,dp[i-1][j-1][1].first)+v[i],v[i]};
            if(dp[i][j][0].first>=inf)    dp[i][j][0]={inf,inf};
            if(dp[i][j][1].first>=inf)    dp[i][j][1]={inf,inf};
        }
    }
    /*for(int i=0;i<n;i++)
    {
        for(int j=0;j<k;j++)
        {
            cout<<dp[i][j].first<<" ";
        }
        cout<<endl;
    }*/
    cout<<min(dp[n-1][k-1][0].first,dp[n-1][k-1][1].first);
}
return 0;
}

• Subtask 1: 0 / 14
• Subtask 2: 0 / 18
• Subtask 3: 0 / 21
• Subtask 4: 0 / 47