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  • Codeforces 854C. Planning

    Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n flights that must depart today, the i-th of them is planned to depart at the i-th minute of the day.

    Metropolis airport is the main transport hub of Metropolia, so it is difficult to keep the schedule intact. This is exactly the case today: because of technical issues, no flights were able to depart during the first kminutes of the day, so now the new departure schedule must be created.

    All n scheduled flights must now depart at different minutes between (k + 1)-th and (k + n)-th, inclusive. However, it's not mandatory for the flights to depart in the same order they were initially scheduled to do so — their order in the new schedule can be different. There is only one restriction: no flight is allowed to depart earlier than it was supposed to depart in the initial schedule.

    Helen knows that each minute of delay of the i-th flight costs airport ci burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.

    Input

    The first line contains two integers n and k (1 ≤ k ≤ n ≤ 300 000), here n is the number of flights, and k is the number of minutes in the beginning of the day that the flights did not depart.

    The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 107), here ci is the cost of delaying the i-th flight for one minute.

    Output

    The first line must contain the minimum possible total cost of delaying the flights.

    The second line must contain n different integers t1, t2, ..., tn (k + 1 ≤ ti ≤ k + n), here ti is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.

    读完题易知肯定让推迟代价最大的航班越早走越好,所以实际就是维护一个优先队列,维护当前能走的最大代价的航班

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    
    #define SIZE 300005
    
    struct node{
      LL pos;
      LL c;
    
      friend bool operator >(const node& lhs,const node& rhs);
    //  friend bool operator <(const node& lhs,const node& rhs);
    };
    
    bool operator >(const node &lhs,const node& rhs){
      if (lhs.c == rhs.c){
        return (lhs.pos < rhs.pos);
      }
      return (lhs.c < rhs.c);
    }
    // bool operator <(const node& lhs,const node& rhs){
    //   return (lhs.c < rhs.c);
    // }
    
    priority_queue<node,vector<node>,greater<node> > pq;
    LL n,k,res = 0;
    LL ans[SIZE] = {0};
    
    int main()
    {
      // freopen("test.in","r",stdin);
      scanf("%I64d %I64d",&n,&k);
    
    
    
      for (int i=1;i<=k+n;i++){
        node temp;
        if (i <= n) {
          LL c;
          scanf("%I64d",&c);
          temp.pos = i; temp.c = c;
          pq.push(temp);
        }
        if (i >= k+1){
          node now = pq.top();
          pq.pop();
          res += now.c * (i - now.pos);
          ans[now.pos] = i;
        }
      }
    
      printf("%I64d
    ",res);
      for (int i=1;i<=n;i++){
        printf("%I64d ",ans[i]);
      }
      return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/ToTOrz/p/7493361.html
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