PAT 1051. Pop Sequence (25)

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最新推荐文章于 2024-04-25 10:28:26 发布

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本文介绍了一个算法问题，即如何判断给定的序列是否为特定条件下的合法弹栈序列。问题中考虑了一个最大容量为M的栈，通过依次压入1到N的整数并随机弹出，探讨了如何验证一个序列是否可以通过这一过程产生。

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1051. Pop Sequence (25)

时间限制

10 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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

提交代码

#include<stdio.h>
#include<stack>
using namespace std;  //使用栈千万别掉了。。
stack <int> S;
int cnt;
bool isOk;
int main()
{	
	int	n,m,k,tmp,i,j;
    //freopen("G:\\in.txt","r",stdin);
	while(scanf("%d%d%d",&n,&m,&k)!=EOF){
		for(i=0;i<k;i++){
			cnt=1;isOk=1;
			while(!S.empty()) S.pop();   //清空栈操作
			for(j=0;j<m;j++){
				scanf("%d",&tmp);
				while(S.empty()||S.top()!=tmp){
					if(S.size()<n&&cnt<=m)  //小于n
						S.push(cnt++);
					else{
						isOk=0;
						break;
					}
				}
				if(!S.empty()&&S.top()==tmp)
					S.pop();
				else
					isOk=0;
			}
			if(isOk==1)
				printf("YES\n");
			else
				printf("NO\n");
		}
	}
    return 0;
}