02-线性结构3 Pop Sequence (25分)

最新推荐文章于 2024-04-05 17:23:19 发布

sysusyf

最新推荐文章于 2024-04-05 17:23:19 发布

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分类专栏： PTA

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该博客探讨了如何判断一个给定的数字序列是否可能是从一个具有特定容量限制的栈中弹出的序列。问题涉及栈操作和序列验证，重点在于理解栈的运作原理和正确推断可能的弹出顺序。

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02-线性结构3 Pop Sequence (25分)

Given a stack which can keep $M$ numbers at most. Push $N$ numbers in the order of 1, 2, 3, ..., $N$ and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if $M$ is 5 and $N$ is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): $M$ (the maximum capacity of the stack), $N$ (the length of push sequence), and $K$ (the number of pop sequences to be checked). Then $K$ lines follow, each contains a pop sequence of $N$ numbers. All the numbers in a line are separated by a space.

Output Specification:

For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.

Sample Input:

5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2

Sample Output:

YES
NO
NO
YES
NO

主要思路：

1、建栈并模拟输出的过程

2、了解栈的特性，下一个将输出数字大于目前已进栈最大数字时，必须依次增大数字并push，直到下一个将输出的数字进栈

3、考虑到无法输出该特定数列的情况（1、进栈时超出容量 2、出栈数字不符合预期）
#include <iostream>
#include <vector>
using namespace  std;
#define Max_Num 1001

typedef struct stack
{
    int Data[Max_Num];
    int Top;
    int Capacity;
}Stack; //顺序存储的栈

void Initialize_Stack(Stack* S,int capacity) //初始化栈，并设定栈的容量
{
    for (int i=0; i<Max_Num; ++i)
    {
        S->Data[i]=0;
    }
    S->Top=0;
    S->Capacity=capacity;
}

bool Is_Full(Stack* S)//判断栈是否已满
{
    if(S->Top==S->Capacity)
        return true;
    else
        return false;
}

bool Is_Empty(Stack* S)//判断栈是否为空
{
    if (S->Top==0)
    {
        return true;
    }else
    {
        return false;
    }
}

bool Push(Stack* S,int value)//入栈操作
{
    if (Is_Full(S))
    {
        return false;
    }
    S->Data[(S->Top)++]=value;
    return true;
}


int Delete(Stack* S)//出栈操作
{
    if (Is_Empty(S))
    {
        return 0;
    }
    return S->Data[--(S->Top)];
}


bool Judge(vector<int> &vec,int M,int N)//判断当前栈容量为M是能否实现长度为N的特定数列的输出
{
    Stack S;
    Initialize_Stack(&S, M);
    int temp=0;
    for (int i=0; i<vec.size(); ++i)
    {
        while(vec[i]>temp)//须输出的数字大于目前已进栈的最大数字，则依次增大并进栈，知道须输出的数字已进栈为止
        {
            if(Push(&S, ++temp)==false)//进栈是超出了栈的容量，则无法输出该特定数列
            {
                return false;
            }
        }
        
        if (Delete(&S)!=vec[i])//出栈的数字不符合预期，则无法输出该特定数列
        {
            return false;
        }
    }
    return true;
}

int main()
{
    //Input
    int M,N,K;
    cin>>M>>N>>K;
    vector<vector<int> > vec(Max_Num);
    int num;
    for (int i=0; i<K; i++)
    {
        for (int k=0; k<N; k++)
        {
            cin>>num;
            vec[i].push_back(num);
        }
    }
    
    //Process
    int i=0;
    while (K--)
    {
        if(Judge(vec[i++],M,N)==true)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    
    
    
    return 0;
}