A1051. Pop Sequence (25)

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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

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <string>
#include <stack> 
using namespace std;


int main(){
    stack<int> st;
    int m,n,k;
    scanf("%d %d %d",&m,&n,&k);
    int a[n];
    while(k--)
    {
        //初始化栈 
        while(!st.empty())
        {
            st.pop();
        } 
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int current=0;
        bool flag=true;
        for(int i=1;i<=n;i++)
        {
            st.push(i);
            if(st.size()>m)
            {
                flag=false;
                break;
            }
            while(!st.empty()&&st.top()==a[current])
            {
                st.pop();
                current++;
            }
        }
        
        if(st.empty()==true&&flag==true)
        printf("YES\n");
        else
        printf("NO\n");
        
    }
    return 0;
}

A1051. Pop Sequence (25)

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原文地址：http://www.cnblogs.com/ligen/p/4313317.html