zoukankan      html  css  js  c++  java
  • 1057 Stack (30 分)分桶法

    1057 Stack (30 分)

    Stack is one of the most fundamental data structures, which is based on the principle of Last In First Out (LIFO). The basic operations include Push (inserting an element onto the top position) and Pop (deleting the top element). Now you are supposed to implement a stack with an extra operation: PeekMedian -- return the median value of all the elements in the stack. With N elements, the median value is defined to be the (N/2)-th smallest element if N is even, or ((N+1)/2)-th if N is odd.

    Input Specification:

    Each input file contains one test case. For each case, the first line contains a positive integer N (105​​). Then N lines follow, each contains a command in one of the following 3 formats:

    Push key
    Pop
    PeekMedian
    

    where key is a positive integer no more than 105​​.

    Output Specification:

    For each Push command, insert key into the stack and output nothing. For each Pop or PeekMedian command, print in a line the corresponding returned value. If the command is invalid, print Invalid instead.

    Sample Input:

    17
    Pop
    PeekMedian
    Push 3
    PeekMedian
    Push 2
    PeekMedian
    Push 1
    PeekMedian
    Pop
    Pop
    Push 5
    Push 4
    PeekMedian
    Pop
    Pop
    Pop
    Pop
    

    Sample Output:

    Invalid
    Invalid
    3
    2
    2
    1
    2
    4
    4
    5
    3
    Invalid
    #include<iostream>
    #include<vector>
    #include<algorithm>
    #include<queue>
    #include<string>
    #include<map>
    #include<set>
    #include<stack>
    #include<string.h>
    #include<cstdio>
    #include<cmath>
    using namespace std;
    int cnt[100005];
    int bucket[100005];
    int Sqrt=113;//常数
    int main()
    {
        int n;
        scanf("%d",&n);
        stack<int>st;
       // st.reserve(n);
        for(int i=0; i<n; i++)
        {
            char ch[30];
            scanf("%s",ch);
            if(ch[1]=='o')
            {
                if(!st.empty())
                {
                    printf("%d
    ",st.top());
                    cnt[st.top()]--;
                    bucket[st.top()/Sqrt]--;
                    st.pop();
                }
                else
                    printf("Invalid
    ");
            }
            else if(ch[1]=='e')
            {
                if(!st.empty())
                {
                    int length=st.size();
                    int m=length%2==0?length/2:(length+1)/2;
                    int id=0,sum=0;
                    while(sum+bucket[id]<m) sum+=bucket[id++];
                    id=id*Sqrt;
                    while(sum+cnt[id]<m) sum+=cnt[id++];
                    printf("%d
    ",id);
                }
                else
                {
                    printf("Invalid
    ");
                }
            }
            else
            {
                int b;
                scanf("%d",&b);
                st.push(b);
                cnt[b]++;
                bucket[b/Sqrt]++;
                //minNum=minNum>b?b:minNum;
            }
        }
        return 0;
    }
     
  • 相关阅读:
    阻止默认事件和冒泡
    js获取元素相对窗口位置
    ios中safari浏览器中date问题
    模拟单选框,多选框
    vue
    js合并两个对象的方法
    oracle 序列
    Oracle生成随机数大全
    JAVA基础面试题
    网速计算
  • 原文地址:https://www.cnblogs.com/zhanghaijie/p/10327136.html
Copyright © 2011-2022 走看看