02-线性结构4 Pop Sequence

02-线性结构4 Pop Sequence

分数 25
作者 陈越
单位 浙江大学

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
1
2
3
4
5
6

Sample Output:

YES
NO
NO
YES
NO
1
2
3
4
5

代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB
C++ (g++)

思路：

定义一个结构体，包含栈顶元素的位置、数组、数组大小。
bool check(vector array,int m,int n)
检查是否能成功出栈
定义一个栈s，用cnt来计数。
s的大小设置为传入的堆栈容量m，栈顶元素置为-1。
开始遍历输入的序列：
栈满的话返回false
否则入栈
当栈顶元素等于队列的队尾值时，出栈，进行上一个队列元素的匹配。

如果遍历完序列 即cnt==n，匹配完所有元素，说明按照这些元素的顺序可以顺利出栈，返回true，否则返回false。

主函数中。输入栈的容量，队列的长度以及需要检查的组数。
然后输入各个队列元素，进行检查。

AC代码：

#include
using namespace std;
#define maxsize 1000

struct Node{
    int Top;//记录堆栈顶元素的位置
    int Array[maxsize];
    int Size;
};

typedef struct Node* Stack;

bool check(vector<int> array,int m,int n)
    //得知道堆栈有多大m，要判断的序列有多大n
{
    Stack s;
    s=new struct Node;
    int cnt=0;
    s->Size=m;
    s->Top=-1;
    for(int i=1;i<=n;i++)
    {
        if(s->Top+1==s->Size) return false;//栈满
        else s->Array[++s->Top]=i;
        while(s->Array[s->Top]==array[cnt])
        {
            s->Top--;//出栈
            cnt++;//数组往后移
        }
    }
    if(cnt==n) return true;
    else return false;
}

int main()
{
    int M,N,K;//M栈 N序列 K组数
    cin>>M>>N>>K;
    vector<int> Array;
    int a;
    for(int i=0;i<K;i++)
    {
        Array.clear();
        for(int j=0;j<N;j++)
        {
            cin>>a;
            Array.push_back(a);
        }
        if(check(Array,M,N)) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}
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