浙大PTA数据结构Pop Sequence

02-线性结构3 Pop Sequence

Given a stack which can keep MMM numbers at most. Push NNN numbers in the order of 1, 2, 3, …, NNN 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 MMM is 5 and NNN 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): MMM (the maximum capacity of the stack), NNN (the length of push sequence), and KKK (the number of pop sequences to be checked). Then KKK lines follow, each contains a pop sequence of NNN 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

#include <iostream>#include <stack>#define Max 1000using namespace std;//给出一段有序数字，判断能否从栈中弹出得到对应的顺序int main(){    stack<int> s;    int i,j,f,x,cnt;    int M,N,K;    int A[Max];    cin>>M>>N>>K;    while(K--)    {        cnt=0;j=1;                                             //cnt为能否成功的标记，每次循环j都是从1开始压栈        for(i=0;i<N;i++) cin>>A[i];        for(i=0;i<N;i++)                                       //从第一个数字遍历        {            if(s.empty())                                      //判断栈顶是否为空             {                    for(;j<=A[i];j++)                          //从当前j元素一直压栈j，j+1,..直到压到A【i】元素                    s.push(j);                    if(s.size()>M) {cnt++;break;}              //判断此时栈长是否超过限制，超过则说明此时得到A【i】元素不合理，即错误                   x=s.top();                                  //x保存即将弹出的栈顶元素                   s.pop();                                    //栈顶元素和A【i】对应，弹出                   continue;             }              if(x>A[i])                                       // 如果之前栈顶元素x即A【i-1】比此时的A【i】元素大，则应该             {                x=s.top();                                     //比较当前栈顶元素与A【i】是否相等，相等则正确，弹出栈顶元素x                if(x==A[i]) s.pop();                else {cnt++;break;}                            //不相等则错误             }             else if(x<A[i])                                   //如果A【i-1】比此时的A【i】元素小，则应该从当前的j开始压栈，j+1，j+2..直到压到A【i】             {                 for(;j<=A[i];j++)                    s.push(j);                if(s.size()>M) {cnt++;break;}                   //如果超过栈的大小限制，则说明不能得到A【i】元素，故退出                 x=s.top();                 s.pop();                 continue;             }        }        if(cnt==1)            cout<<"NO"<<endl;        else cout<<"YES"<<endl;        while(!s.empty())                                      //每次循环后，要清空栈，防止栈中元素对下次循环的影响                s.pop();    }    return 0;}