zoukankan      html  css  js  c++  java
  • P

    P - Psychos in a Line
    Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
    Submit Status
    Appoint description: 

    Description

    There are n psychos standing in a line. Each psycho is assigned a unique integer from 1 to n. At each step every psycho who has an id greater than the psycho to his right (if exists) kills his right neighbor in the line. Note that a psycho might kill and get killed at the same step.

    You're given the initial arrangement of the psychos in the line. Calculate how many steps are needed to the moment of time such, that nobody kills his neighbor after that moment. Look notes to understand the statement more precise.

    Input

    The first line of input contains integer n denoting the number of psychos, (1 ≤ n ≤ 105). In the second line there will be a list of nspace separated distinct integers each in range 1 to n, inclusive — ids of the psychos in the line from left to right.

    Output

    Print the number of steps, so that the line remains the same afterward.

    Sample Input

    Input
    10
    10 9 7 8 6 5 3 4 2 1
    Output
    2
    Input
    6
    1 2 3 4 5 6
    Output
    0
     
    const int INF = 1000000000;
    const double eps = 1e-8;
    const int maxn = 301000;
    typedef pair<int,int> ii;
    int a[maxn];
    int next[maxn];//链表指针
    int main() 
    {
        //freopen("in.txt","r",stdin);
        int n;
        while(cin>>n)
        {
            queue<ii> q;
            rep(i,0,n)
            {
                scanf("%d",&a[i]);
                next[i] = i+1;
            }
            
            repd(i,n-2,0)
            {
                if(a[i] > a[i+1])
                    q.push(ii(i,0));
            }
            
            int ans = 0;
            while(!q.empty())
            {
                int pos = q.front().first;
                int cas = q.front().second;
                ans = max(ans,cas);
                q.pop();
                if(next[pos] != n && a[pos] > a[next[pos]])
                {
                    q.push(ii(pos,cas+1));
                    next[pos] = next[next[pos]];
                }
            }
            
            cout<<ans<<endl;
        }
        return 0;
    }
  • 相关阅读:
    无法设置 / 添加网络打印机?报错 无法保持设置?
    tp剩余未验证内容-5
    再谈 iptables 防火墙的 指令配置
    tp剩余未验证内容-4
    tp剩余未验证内容-3
    CentOS7.4安装配置mysql8 TAR免安装版
    CentOS7中systemctl的使用与CentOS6中service的区别
    CentOS下如何查看并杀死僵尸进程
    CentOS SVN服务器管理多项目
    swoole+Redis实现实时数据推送
  • 原文地址:https://www.cnblogs.com/DreamHighWithMe/p/3427926.html
Copyright © 2011-2022 走看看