Pop Sequence (栈)
Pop Sequence (栈)
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
一个数要出栈，那么比这个数小的数一定要先进栈
设 tem 为进栈的数。如果 栈为空 或 出栈的数 不等于 栈顶数，那么 tem进栈 后 自增。如果 出栈的数 等于 栈顶数，则出栈。如果栈溢出，那就错误（可能是要出栈的数太大 或者是 要出栈的数 小于 此刻的栈顶元素 ）
#include <iostream>
#include <string>
#include<vector>
#include <stack>
using namespace std;
int main()
{
int i,j,m,k,n,x;
while(cin>>m)
{
cin>>n>>k;
for(i=0;i<k;i++)
{
bool is=true;
stack<int> tt;
int tem=1;
for(j=0;j<n;j++)
{
cin>>x;
while(tt.size()<=m && is)
{
if(tt.empty() || tt.top()!=x)
tt.push(tem++);
else if(tt.top()==x)
{
tt.pop();
break;
}
}
if(tt.size() > m)
{
is=false;
}
}
if(is) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}
return 0;
}