题目链接
题目大意：有一个邮差，邮局在坐标0点，接下来有n次请求，每次请求包括一个坐标点和要送的信件数目，邮差所能带的信件数目最大为k，每次送完后要回到0点，最后也要回到0点，求运送完成的最小时间。
贪心，先走最远的，一开始写的暴力跑，TLE了，可以选择把最远的可以优化的容量都累计到相邻的请求中去，这样只需要跑一遍for循环即可。
AC代码：

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
struct node
{
    long long
     x,m;
} a[2000],b[2000];
int cmp(node a,node b)
{
    return a.x<b.x;
}
int cmp2(node a,node b)
{
    return a.x>b.x;
}
int main()
{
    int n,k;
    cin>>n>>k;
    node tmp;
    int top1 =0 ;
    int top2 =0 ;
    for(int i=1; i<=n; i++)
    {
        scanf("%lld%lld",&tmp.x,&tmp.m);
        if(tmp.x<0)a[++top1] = tmp;
        else if(tmp.x>0)b[++top2] = tmp;

    }
    long long ans = 0;
    sort(a+1,a+1+top1,cmp);
    sort(b+1,b+1+top2,cmp2);
    int kk = 0;
     for(int i=1; i<=top1; i++)
    {
      // cout<<b[i].m<<"*"<<endl;
      if(a[i].m<0)
      {
        if(i>top1-1)continue;
        else
        {
            a[i+1].m += a[i].m;
        }
      }
      else
      {
          int hh = (a[i].m+k-1)/k;
          ans+=2*(hh)*abs(a[i].x);
          a[i].m = a[i].m%k;
          if(i<top1&&a[i].m)
          a[i+1].m-=(k-a[i].m);
      }
    }
  // cout<<ans<<endl;

    for(int i=1; i<=top2; i++)
    {
      // cout<<b[i].m<<"*"<<endl;
      if(b[i].m<0)
      {
        if(i>top2-1)break;
        else
        {
            b[i+1].m += b[i].m;
        }
      }
      else
      {
          int hh = (b[i].m+k-1)/k;
          ans+=2*(hh)*abs(b[i].x);
          b[i].m = b[i].m%k;
          if(i<top2&&b[i].m)
          b[i+1].m-=(k-b[i].m);
      }
    }
    cout<<ans<<endl;
    return 0;
}

TLE代码：

#include <iostream>
#include<bits/stdc++.h>
using namespace std;
struct node
{
    long long
     x,m;
} a[2000],b[2000];
int cmp(node a,node b)
{
    return a.x<b.x;
}
int cmp2(node a,node b)
{
    return a.x>b.x;
}
int main()
{
    int n,k;
    cin>>n>>k;
    node tmp;
    int top1 =0 ;
    int top2 =0 ;
    for(int i=1; i<=n; i++)
    {
        scanf("%lld%lld",&tmp.x,&tmp.m);
        if(tmp.x<0)a[++top1] = tmp;
        else if(tmp.x>0)b[++top2] = tmp;

    }
    long long ans = 0;
    sort(a+1,a+1+top1,cmp);
    sort(b+1,b+1+top2,cmp2);
    int kk = 0;
    for(int i=1; i<=top1; i++)
    {
        if(a[i].m<=0)continue;
        double gg = a[i].m*1.0/k;
        long long u = ceil(gg);
        kk = k;
        if(a[i].m>=k)
        {

            kk-=a[i].m%k;
            ans+=(2*u*abs(b[i].x));

        }
        else
        {
            ans+=(2*abs(a[i].x));
             kk-=a[i].m;

        }
        if(a[i].m%k==0)continue;
        while(kk>0)
        {
            int j = i+1;
            if(kk<a[j].m)
            {
                a[j].m-=kk;
                kk= 0;
                break;
            }
            else
            {
                a[j].m = 0;
                kk-=a[j].m;
            }
            j++;
            if(j>top1)break;
            //cout<<kk<<endl;
        }

    }
    //cout<<ans<<endl;

    for(int i=1; i<=top2; i++)
    {
      // cout<<b[i].m<<"*"<<endl;
       if(b[i].m<=0)continue;
       double gg = b[i].m*1.0/k;
        long long u = ceil(gg);
        kk = k;
        if(b[i].m>=k)
        {
             kk-=b[i].m%k;
            ans+=(2*u*abs(b[i].x));
        }
        else
        {
            kk-=b[i].m;
            ans+=(2*abs(b[i].x));
        }
        if(b[i].m%k==0)continue;

        while(kk>0)
        {
            int j = i+1;
            if(kk<b[j].m)
            {
                b[j].m-=kk;
                kk= 0;
                break;
            }
            else
            {
                b[j].m = 0;
                kk-=b[j].m;
            }
            j++;
            if(j>top2)break;
          // cout<<kk<<endl;
        }
    }
    cout<<ans<<endl;
    return 0;
}