02-线性结构3. Pop Sequence

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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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

简单的模拟堆栈的活动，判断YES or NO。作为堆栈的练习。用数组和结构体皆可，数组简单一些。所以在这里传一个结构体做CODE。
#include<stdio.h>#include<stdlib.h>int I = 0;typedef struct stack{int a;struct stack* next;}Stack;Stack* Free(Stack* p);int pop(Stack*p); int IsFull(Stack* p);Stack* push(Stack* p,int value);int main(){int m,n,k;int i,j,t = 1;scanf("%d %d %d",&m,&n,&k);int list[k][n];for(i = 0;i<k;i++)for(j = 0;j<n;j++)scanf("%d",&list[i][j]);//创建堆栈 Stack *SP;SP = (Stack*)malloc(sizeof(Stack));SP->next = NULL;SP->a = 0;//进出栈 for(i = 0;i<k;Free(SP),t = 1,i++){for(j = 0;j<n;j++){int Pi = list[i][j];if(IsFull(SP)){push(SP,t);t++;}if(!IsFull(SP)&&Pi<SP->next->a){printf("NO\n");break;}while(!IsFull(SP)&&Pi>SP->next->a){push(SP,t);t++;if(I>m){printf("NO\n");goto out;}if(t>n)t = 1;}if(!IsFull(SP)&&Pi == SP->next->a){pop(SP);if(j==n-1)printf("YES\n");}}out:;}return 0;}int pop(Stack*p){int t;Stack* New;if(IsFull(p)){printf("\nkong\n");return 0;}else{New = p->next;p->next = New->next; t = New->a;free(New);I--;return t;}}int IsFull(Stack* p){return (p -> next == NULL);}Stack* push(Stack* p,int value){I++;Stack *New;New = (Stack*)malloc(sizeof(struct stack));New->a = value;New->next = p->next;p->next = New;return p;} Stack* Free(Stack* p){while(p->next!=NULL)pop(p); return p;}

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