1051 Pop Sequence （25 分）模拟堆栈

题目

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

解题思路

　　题目大意： 有一个栈，大小为M，数字1~N依次入栈，给出K个出栈序列，判断该序列是否满足出栈要求。
　　解题思路： 入栈顺序是确定的，我们只需要模拟这个过程，从1到N入栈，如果遇到题目所给序列，那么即刻出栈，遍历到最后的结果应该是空栈空序列，刚好满足。

/*
** @Brief:No.1051 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-12-21
** @Solution: https://blog.csdn.net/CV_Jason/article/details/85166926
*/
#include
#include 
#include

using namespace std;

int main(){
	int M,N,K;
	while(cin>>M>>N>>K){
		vector<int> seq(N);
		for(int i=0;i<K;i++){
			for(int j=0;j<N;j++){
				cin>>seq[j];
			}
			bool isStack = true;
			stack<int> st;
			int index = 0;
			for(int j=1;j<=N;j++){
				st.push(j);
				if(st.size()>M){
					isStack = false;
					break;
				}
				while(!st.empty()&&st.top()==seq[index]){
					st.pop();
					index++;
				}
			}
			if(!st.empty()||index!=seq.size()){
				isStack = false;
			}
			cout<<(isStack?"YES":"NO")<<endl;
		}
	}
	return 0;
}

总结

　　当然，也可以不用stl提供的stack，这道题比较简单，我们可以用手动实现一个stack，在时间和空间上，都是高效的。但使用top指针指向stack的时候要注意，预留一个-1作为栈底标志，那么入栈的时候应该是++top而不是top++。

/*
** @Brief:No.1051 of PAT advanced level.
** @Author:Jason.Lee
** @Date:2018-12-21
** @Solution: Accepted!
*/
#include
using namespace std;
int stack[1000];

int main(){
	int M,N,K;
	while(cin>>M>>N>>K){
		for(int i=0;i<K;i++){
			int seq[N];
			for(int j=0;j<N;j++){
				cin>>seq[j];
			}
			int top = -1;
			int index = 0;
			bool isStack = true;
			for(int j=1;j<=N;j++){
				stack[++top] = j;
				if(top>=M){
					isStack = false;
					break; 
				}
				while(top!=-1&&stack[top]==seq[index]){
					top--;
					index++;
				}
			}
			
			if(top!=-1||index!=N){
				isStack = false;
			}
			cout<<(isStack?"YES":"NO")<<endl;
		}
	}
}