1051 Pop Sequence （25 分）简单模拟

1051 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
思路：
　　定义一个栈，进行出栈模拟。
　　　　1、循环下标从1开始，因为进栈序列是从1，开始的。
　　　　2、如果当前栈中元素个数大于最大值m则返回false
　　　　3、如果栈顶元素和出栈序列第一个元素相同，则循环出栈，直到栈顶元素和出栈序列元素不匹配
　　　　4、循环结束后，如果栈中还有元素则返回false
　　　　5、否则，返回true
　　详细请看代码
　　

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<string>
#include<map>
#include<set>
#include<stack>
using namespace std;

bool check(int a[],int n,int m)
{
    stack<int> st;
    int top=1;
    for(int i=1; i<n+1; i++)
    {
        st.push(i);
        if(st.size()>m)
            return false;
        //如果栈顶元素等于出栈序列第一个则出栈
        while(!st.empty()&&st.top()==a[top])
        {
            st.pop();
            top++;
        }


    }
    if(!st.empty())
        return false;
    return true;
}


int main()
{
    int m,n,k;
    cin>>m>>n>>k;
    int a[n+1];
    for(int i=0; i<k; i++)
    {
        for(int j=1; j<n+1; j++)
            cin>>a[j];
        if(check(a,n,m))
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}