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;
}