数据结构 邓俊辉 PA#2 列车调度(Train) 题解

Description

Figure 1 shows the structure of a station for train dispatching.

Figure 1

In this station, A is the entrance for each train and B is the exit. S is the transfer end. All single tracks are one-way, which means that the train can enter the station from A to S, and pull out from S to B. Note that the overtaking is not allowed. Because the compartments can reside in S, the order that they pull out at B may differ from that they enter at A. However, because of the limited capacity of S, no more that m compartments can reside at S simultaneously.

Assume that a train consist of n compartments labeled {1, 2, …, n}. A dispatcher wants to know whether these compartments can pull out at B in the order of {a1, a2, …, an} (a sequence). If can, in what order he should operate it?

Input

Two lines:

1st line: two integers n and m;

2nd line: n integers separated by spaces, which is a permutation of {1, 2, …, n}. This is a compartment sequence that is to be judged regarding the feasibility.

Output

If the sequence is feasible, output the sequence. “Push” means one compartment goes from A to S, while “pop” means one compartment goes from S to B. Each operation takes up one line.

If the sequence is infeasible, output a “no”.

Example 1

Input

1
2
5 2
1 2 3 5 4

Output

1
2
3
4
5
6
7
8
9
10
push
pop
push
pop
push
pop
push
push
pop
pop

Example 2

Input

1
2
5 5
3 1 2 4 5

Output

1
No

Restrictions

1 <= n <= 1,600,000

0 <= m <= 1,600,000

Time: 2 sec

Memory: 256 MB

描述

某列车调度站的铁道联接结构如Figure 1所示。

其中,A为入口,B为出口,S为中转盲端。所有铁道均为单轨单向式:列车行驶的方向只能是从A到S,再从S到B;另外,不允许超车。因为车厢可在S中驻留,所以它们从B端驶出的次序,可能与从A端驶入的次序不同。不过S的容量有限,同时驻留的车厢不得超过m节。

设某列车由编号依次为{1, 2, …, n}的n节车厢组成。调度员希望知道,按照以上交通规则,这些车厢能否以{a1, a2, …, an}的次序,重新排列后从B端驶出。如果可行,应该以怎样

的次序操作?

输入

共两行。

第一行为两个整数n,m。

第二行为以空格分隔的n个整数,保证为{1, 2, …, n}的一个排列,表示待判断可行性的驶出序列{a1,a2,…,an}。

输出

若驶出序列可行,则输出操作序列,其中push表示车厢从A进入S,pop表示车厢从S进入B,每个操作占一行。

若不可行,则输出No。

样例

见英文题面

限制

1 ≤ n ≤ 1,600,000

0 ≤ m ≤ 1,600,000

时间:2 sec

空间:256 MB

模拟栈混洗:(自写Stack,因为该OJ禁止STL)

代码:

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#include<iostream>
#include<cstdio>
#include<cstring>
template <typename T> struct Stack{
    private:
        int _capacity;
        int _size;
        T * array;
    public:
        Stack(int capacity){
            this->array=new T[capacity];
            this->_capacity=capacity;
            this->_size=0;
        }
        ~Stack(){
            delete [] array;
        }
        bool empty(){
            return (_size==0);
        }
        bool full(){
            return (_size==_capacity);
        }
        int size(){
            return _size;
        }
        void push(T content){
            this->array[_size]=content;
            _size++;
        }
        void pop(){
            _size--;
        }
        T & top(){
            return array[_size-1];
        }
};
int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    Stack<int> S(m);
    bool * record=new bool[2*n+1];
    int * array=new int[n];
    for(int i=0;i<n;i++)scanf("%d",array+i);
    int arrptr=0,recordptr=0;
    for(int i=1;i<=n;i++){
        if(!S.full()){
            S.push(i);
            record[recordptr++]=true;
            while(!S.empty()&&array[arrptr]==S.top()){
                S.pop();
                record[recordptr++]=false;
                arrptr++;
            }
        }else{
            printf("No");
            return 0;
        }
    }
    if(!S.empty()){
        printf("No");
        return 0;
    }else{
        for(int i=0;i<recordptr;i++){
            if(record[i])printf("push\n");
            else printf("pop\n");
        }
        return 0;
    }
}

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