PAT A1051 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

实现思路：

考察给一个序列是否为一个合法的出栈序列，这里只需要模拟栈的出栈入栈操作即可。

AC代码：

#include <iostream>
#include <stack>
#include <vector>
using namespace std;
const int MAXN=1001;
vector<int> sq[MAXN];
int n,m,k,val;

bool testRes(vector<int> &q) {
	stack<int> st;
	int cnt=1;
	for(int i=0; i<q.size(); i++) {
		if(cnt<=q[i]) {
			while(cnt<=q[i]) {
				st.push(cnt);
				if(st.size()>m) return false;
				if(cnt==q[i]) st.pop();
				cnt++;
			}
		} else {
			if(st.size()>0&&st.top()==q[i]) st.pop();
			else return false;
		}
	}
	return true;
}

int main() {
	cin>>m>>n>>k;
	for(int i=0; i<k; i++) {
		for(int j=0; j<n; j++) {
			scanf("%d",&val);
			sq[i].push_back(val);
		}
		if(testRes(sq[i])) printf("YES");
		else printf("NO");
		printf("\n");
	}
	return 0;
}