zoukankan      html  css  js  c++  java
  • Codeforces 709B 模拟

    B. Checkpoints
    time limit per test:1 second
    memory limit per test:256 megabytes
    input:standard input
    output:standard output

    Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x1, x2, ..., xn. Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.

    Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.

    Input

    The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000,  - 1 000 000 ≤ a ≤ 1 000 000) — the number of checkpoints and Vasya's starting position respectively.

    The second line contains n integers x1, x2, ..., xn ( - 1 000 000 ≤ xi ≤ 1 000 000) — coordinates of the checkpoints.

    Output

    Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.

    Examples
    Input
    3 10
    1 7 12
    Output
    7
    Input
    2 0
    11 -10
    Output
    10
    Input
    5 0
    0 0 1000 0 0
    Output
    0
    Note

    In the first sample Vasya has to visit at least two checkpoints. The optimal way to achieve this is the walk to the third checkpoints (distance is 12 - 10 = 2) and then proceed to the second one (distance is 12 - 7 = 5). The total distance is equal to 2 + 5 = 7.

    In the second sample it's enough to visit only one checkpoint so Vasya should just walk to the point  - 10.

    题目连接:http://codeforces.com/contest/709/problem/B


    题意:一条直线上面有n个标记,一个起点a。从起点开始出发最少经过n-1个标记最少需要的路程。
    思路:模拟。路程最少,所以只需要经过n-1个点。a的右边经过cou个点,则坐标经过n-1-cou个点。
    代码:
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    int x[100100];
    int l[100100],r[100100];
    int main()
    {
        int i,n,a;
        scanf("%d%d",&n,&a);
        int cou=0;
        for(i=1; i<=n; i++)
        {
            scanf("%d",&x[i]);
            if(x[i]<=a) cou++;
        }
        int cou1=cou;
        sort(x+1,x+n+1);
        for(i=1; i<=n; i++)
        {
            if(x[i]<=a) l[cou]=x[i],cou--;
            else r[++cou]=x[i];
        }
        l[0]=r[0]=a;
        int cou2=cou;
        int ans=10000000;
        for(i=0; i<=cou2; i++)
        {
            if(i>n-1) break;
            if(n-1-i>cou1) continue;
            int sign;
            if(i==0) sign=a-l[n-1];
            else if(i==n-1) sign=r[n-1]-a;
            else
            {
                if(r[i]-a<=a-l[n-1-i]) sign=r[i]-a+r[i]-l[n-1-i];
                else sign=a-l[n-1-i]+r[i]-l[n-1-i];
            }
            if(sign<ans) ans=sign;
        }
        cout<<ans<<endl;
        return 0;
    }
    View Code
    I am a slow walker,but I never walk backwards.
  • 相关阅读:
    Elastic的should + bool 多字段完全匹配或查找
    MySQL慢日志
    Elastic的IN查找
    Elastic的字符串查找
    JavaScript获取当前时间戳
    原码, 反码, 补码学习笔记
    PHP渲染压缩HTML
    JavaScript的深拷贝
    JavaScript的变量的let const var的区别
    关于一个值该不该default null的判定
  • 原文地址:https://www.cnblogs.com/GeekZRF/p/5866133.html
Copyright © 2011-2022 走看看