Submission #60828

#	Time	Username	Problem	Language	Result	Execution time	Memory
60828		dukati8	Boxes with souvenirs (IOI15_boxes)	C++14	20 / 100	2 ms	432 KiB

boxes

#include "boxes.h"
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a; i<int(b); i++)
#define all(x) x.begin(),x.end()

long long delivery(int N, int K, int L, int p[]) {
    if (K==1) {
      long long tot=0;
      rep(i,0,N) {
        tot+=2*min(p[i],L-p[i]);
      }
      return tot;
    }
    map<long long,int> places;
    long long biggestplace=0;
    rep(i,0,N) {
      long long curr=p[i];
      biggestplace=max(biggestplace,curr);
      if (places.find(curr)!=places.begin()) places[curr]++;
      else places[curr]=1;
    }

    map<int,long long> dpthing; //This dp should be, for every possible amount of boxes that chould have been transported
    //to all the teams before and including the place we are at, minimum time to get there with those boxes
    dpthing.insert(make_pair(0,0));
    long long prevplace=0;
    long long extratime=0;
    int need=0;
    for (auto a:places) {
      long long loc=a.first;
      int amount=a.second;
      extratime+=(amount/K)*8*min(loc,L-loc);
      amount%=K;
      if (amount==0) continue;
      long long mintime=1000000000000000;
      need+=amount;
      vector<int> rem;
      for (auto b:dpthing) {
        int have=b.first;
        mintime=min(mintime,b.second);
        if (have>=need) break;
        rem.push_back(have);
      }
      for (auto b:rem) {
        dpthing.erase(b);
      }
      //Now everything in dpthing has enough for this new location, they move there and increase time by the distance
      extratime+=min(loc-prevplace,L-loc+prevplace);
      //We can also go back and fetch new boxes, this should be added to the previous smallest mintime
      mintime+=min(prevplace,L-prevplace) + min(loc,L-loc); //Time to go back and fetch more and go to new place
      mintime-=min(loc-prevplace,L-loc+prevplace); //This was added to extratime instead
      dpthing.insert(make_pair(need-amount+K,mintime)); //We would have need-amount+K after that
      prevplace=loc;
    }
    //Now we take the least value in dpthing, add time to get back and extratime to it
    long long best=1000000000000000;
    for (auto a:dpthing) {
      best=min(best,a.second);
    }
    return best + min(biggestplace,L-biggestplace) + extratime;

}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30