office key （重要结论）

2022-03-06 19:04:34 阅读：198 来源： 互联网

标签：结论 ch office point people located key

重要结论：钥匙一定是连续的。

A. Office Keys
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
There are n people and k keys on a straight line. Every person wants to get to the office which is located on the line as well. To do that, he needs to reach some point with a key, take the key and then go to the office. Once a key is taken by somebody, it couldn't be taken by anybody else.

You are to determine the minimum time needed for all n people to get to the office with keys. Assume that people move a unit distance per 1 second. If two people reach a key at the same time, only one of them can take the key. A person can pass through a point with a key without taking it.

Input
The first line contains three integers n, k and p (1 ≤ n ≤ 1 000, n ≤ k ≤ 2 000, 1 ≤ p ≤ 109) — the number of people, the number of keys and the office location.

The second line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ 109) — positions in which people are located initially. The positions are given in arbitrary order.

The third line contains k distinct integers b1, b2, ..., bk (1 ≤ bj ≤ 109) — positions of the keys. The positions are given in arbitrary order.

Note that there can't be more than one person or more than one key in the same point. A person and a key can be located in the same point.

Output
Print the minimum time (in seconds) needed for all n to reach the office with keys.

Examples
inputCopy
2 4 50
20 100
60 10 40 80
outputCopy
50
inputCopy
1 2 10
11
15 7
outputCopy
7
Note
In the first example the person located at point 20 should take the key located at point 40 and go with it to the office located at point 50. He spends 30 seconds. The person located at point 100 can take the key located at point 80 and go to the office with it. He spends 50 seconds. Thus, after 50 seconds everybody is in office with keys.

View problem

#include <bits/stdc++.h>
using namespace std;
#define ri register int
#define M 10005


template  <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
}

int n,m,md;
int p[M],a[M];
int main(){
    
    read(n);read(m);read(md);
    for(ri i=1;i<=n;i++)
    read(p[i]);
    for(ri j=1;j<=m;j++)
    read(a[j]);
    
    sort(p+1,p+1+n);
    sort(a+1,a+1+m);
    
    long long ans=1e18;
    for(ri i=1;i<=(m-n+1);i++)
    {
        long long tmp=0;
        for(ri j=1;j<=n;j++)
        {
          long long  aa=abs(p[j]-a[i+j-1])+abs(a[i+j-1]-md);tmp=max(tmp,aa);
        }
        ans=min(ans,tmp);
    }
    printf("%lld",ans);
    return 0;
    
    
}

View Code

标签：结论,ch,office,point,people,located,key
来源： https://www.cnblogs.com/Lamboofhome/p/15973003.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。

ICode9

office key （重要结论）