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
题意：
给定一个有固定容量的栈，1,2,...,n是入栈序列，元素出栈顺序随意，现给定出栈顺序(e.g.1~n的一个排列)，问这个出栈顺序是否合理，合理输出"YES"，否则输出"NO"。
题解：
开一个队列，开一个栈，输入一个数，就队列中这个数及之前的数放入栈中，放入不能超过容量。然后看栈顶元素是不是这个数，是就下一个，不是就标记NO。
AC代码：
#include<iostream>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
#include<vector>
#include<string>
#include<cstring>
using namespace std;
int n,m,k;
stack<int>s;
queue<int>q;
int main(){
cin>>n>>m>>k;
while(k--){
while(!s.empty()) s.pop();
while(!q.empty()) q.pop();
int f=;
for(int i=;i<=m;i++) q.push(i);
for(int i=;i<=m;i++){
int x;
cin>>x;
if(s.empty()||s.top()!=x){
while(!q.empty()){
if(s.size()<n) {
//cout<<"把"<<q.front()<<"放栈"<<endl;
s.push(q.front());
q.pop();
}
else break;
if(s.top()==x) break;
}
}
if(s.top()==x){
s.pop();
//cout<<x<<"踢出栈"<<endl;
continue;
}
f=;
}
if(f) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return ;
}