Postman

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4096

题意：x=0的位置是邮局，邮递员每次可以带k封信，需要回邮局拿信，最后送完信后不必回邮局，求最小路程

C++版本一

题解：

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define FI first
#define SE second
#define PB push_back
typedef pair<int,LL>PII;
const int N=2e5+7;
const LL INF=1e17,mod=998244353;
int n,m,k;
LL a[N],b[N];
LL ans;
int main()
{
    int t;
    cin>>t;
    while(t--){
            ans=0;
        scanf("%d%d",&n,&k);
        LL x;
        int cnt=0,cot=0;
        for(int i=1;i<=n;i++){
            scanf("%lld",&x);
            if(x==0)continue;
            if(x<0)a[++cnt]=-x;
            else b[++cot]=x;
        }
        sort(a+1,a+cnt+1);
        sort(b+1,b+cot+1);
        if(a[cnt]>b[cot]){
            ans+=a[cnt];
            cnt=max(cnt-k,0);
        }
        else {
            ans+=b[cot];
            cot=max(cot-k,0);
        }
        //cout<<ans<<endl;
        while(cnt>0||cot>0){
            if(a[cnt]>b[cot]){
                ans+=a[cnt]*2;
                cnt=max(cnt-k,0);
            }
            else{
                ans+=b[cot]*2;
                cot=max(cot-k,0);
            }
        }
        printf("%lld\n",ans);
    }
	return 0;
}

C++版本二

题解：

1、从两边往中间走；

2、减掉一个绝对值最大的点；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
ll c[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    srand((unsigned)time(NULL));
    scanf("%d",&t);
    //t=rand()%(2000)+10000;
    while(t--){
        scanf("%d%d",&n,&k);
        ll maxl=0;
        for(int i=1;i<=n;i++){
            scanf("%lld",&c[i]);
            maxl=max(maxl,abs(c[i]));
        }
        sort(c+1,c+n+1);
        int l=0;
        int r=n+1;
        ans=0;
        for(int i=n;i>=1;i--){
            if(c[i]>0){
                r=i;
            }else{
                break;
            }
        }
        for(int i=1;i<=n;i++){
            if(c[i]<0){
                l=i;
            }else{
                break;
            }
        }
 
        for(int i=n;i>=r;i-=k){
            ans+=2ll*abs(c[i]);
        }
        for(int i=1;i<=l;i+=k){
            ans+=2ll*abs(c[i]);
        }
        cout<<ans-maxl<<endl;
    }
 
#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}