02-线性结构4 Pop Sequence

02-线性结构4 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

#include <iostream>#include <malloc.h>using namespace std;typedef struct node{    int data;    struct node *next;}linkStack;linkStack *createStack(){    linkStack *s;    s=(linkStack *)malloc(sizeof(linkStack));    s->next=NULL;    return s;}int isEmpty(linkStack * s){    return s->next==NULL;}void push(int item,linkStack *s){    struct node *temCell;    temCell=(linkStack *)malloc(sizeof(linkStack));    temCell->data=item;    temCell->next=s->next;    s->next=temCell;}int pop(linkStack *s){    if(isEmpty(s)) return NULL;    else{        int data;        linkStack *first;        first=(linkStack *)malloc(sizeof(linkStack));        first=s->next;        s->next=first->next;        data=first->data;        free(first);        return data;    }}int main(){    int m,n,k,num;    cin>>m>>n>>k;    int seques[n+1]={1};    for(int i=0;i<k;i++){        int j,cnt=0,isOrno=1;        linkStack *s;        s=createStack();        for(j=0;j<n;j++){            cin>>num;            if(!isEmpty(s) && num<s->next->data){                isOrno=0;                break;            }            else{                for(int i=1;i<=num;i++){                    if(seques[i]){                        push(i,s);                        seques[i]=0;                        cnt++;                    }                }            }            if(cnt>m){                isOrno=0;                break;            }            pop(s);            seques[num]=0;            cnt--;        }        if(isOrno) cout<<"YES"<<endl;        else cout<<"NO"<<endl;        while(j+1<n){            cin>>num;            j++;        }    }    return 0;}