PTA 02-线性结构4 Pop Sequence (25分)

题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/665

5-3 Pop Sequence   (25分)

Given a stack which can keep MM numbers at most. Push NN numbers in the order of 1, 2, 3, ..., NN 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 MM is 5 and NN 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): MM (the maximum capacity of the stack), NN (the length of push sequence), and KK (the number of pop sequences to be checked). Then KK lines follow, each contains a pop sequence of NN 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
/*
评测结果
时间	结果	得分	题目	编译器	用时(ms)	内存(MB)	用户
2017-07-08 15:55	答案正确	25	5-3	gcc	3	1	
测试点结果
测试点	结果	得分/满分	用时(ms)	内存(MB)
测试点1	答案正确	15/15	2	1
测试点2	答案正确	3/3	2	1
测试点3	答案正确	2/2	2	1
测试点4	答案正确	2/2	3	1
测试点5	答案正确	1/1	2	1
测试点6	答案正确	2/2	1	1

*/
#include<stdio.h>
#define MAXLEN 1000

/*下面这段代码只能拿18/25分 
int main()
{
	int i,j,m,n,k,last,flag,temp;
	scanf("%d %d %d",&m,&n,&k);
	for (i=0;i<k;i++){
		flag=1;
		last=0;
		for(j=0;j<n;j++){
			scanf("%d",&temp);
			if (temp-last>m) flag=0;
			last=temp;
		}
		if(flag==0) printf("NO
");
		else printf("YES
");
			
	}
}
*/

struct stack{
	int data[MAXLEN];
	int top;
	int max;
};

struct stack workstack;


int GetTop(struct stack stc)
{
	if(stc.top>=0) return stc.data[stc.top];
	else return 0;
}

int Push(struct stack *stc,int item)
{
	if (stc->top==MAXLEN-1) return 0;
	else {
		stc->data[++(stc->top)]=item;
//		printf("--PUSH %d,%d
",stc->data[(stc->top)],stc->top);//test
		return 1;
	}
}

int Pop(struct stack *stc)
{
	if(stc->top<0) return 0;
	else{
//		printf("--POP %d,%d
",stc->data[(stc->top)],stc->top);//test
		return stc->data[(stc->top)--];
	}
}

int main()
{	
	int i,j,m,n,k,numforpush,errorflag,temp;
	scanf("%d %d %d",&m,&n,&k);
	workstack.max=m;
	for (i=0;i<k;i++){
		errorflag=0;
		numforpush=1;
		workstack.top=-1;
		for(j=0;j<n;j++){
			scanf("%d",&temp);
			if (errorflag==1) continue;
//			printf("--GET %d
",temp);//test
			while(workstack.top<workstack.max-1 && numforpush<=n && GetTop(workstack)!=temp){
//				printf("--IN WHILE workstack.top=%d,max=%d,numforpush=%d
",workstack.top,workstack.max,numforpush);//test
				Push(&workstack,numforpush);
				numforpush++;
			}
			
			if(GetTop(workstack)==temp){
				Pop(&workstack);
			}
			else{
				errorflag=1;
			}
			
		}
		if(errorflag==1)
			printf("NO");
		else
			printf("YES");
		if(i!=k-1) putchar('
');
	}
	
	
}


原文地址:https://www.cnblogs.com/gk2017/p/7140524.html