1051 Pop Sequence (25)

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

思路：

有个容量限制为m的栈，分别把1，2，3，…，n入栈,给出一个系列出栈顺序，问这些出栈顺序是否可能.模拟出栈

C++：

#include "stack"
#include "cstdio"
#include "iostream"
#include "vector"
using namespace std;
int main(){
	int n,m,k;
	scanf("%d %d %d",&m,&n,&k);
	for(int i = 0; i < k; i++) {
		bool flag = false;
		stack<int> s;
		vector<int> v(n + 1);
		for(int j = 1; j <= n; j++)
			scanf("%d", &v[j]);
		int current = 1;
		for(int j = 1; j <= n; j++) {//按照1、2、3......n的序列进入栈
			s.push(j);
			if(s.size() > m) break;//超过栈的容量限制
			while(!s.empty() && s.top() == v[current]) {//栈顶元素若与序列相同则一直弹出
				s.pop();//弹出
				current++;
			}
		}
		if(current == n + 1) flag = true;//该序列可以
		if(flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}