PAT1051:Pop Sequence

1051. Pop Sequence (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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
栈的应用。这里直接用vector模拟栈，不得不说vector太强大了，既可以当栈又可以当队列用。

这道题需要理清楚弹出序列不满足的条件：
1.压进去数字后的栈容量大于限定的容量。
2.弹出序列需求的弹出元素不在栈顶。

简单过程：
1.用一个bool值isSuccess来标识弹出序列是否可行，默认为true。
2.保存第一次压入操作的最大元素cur，然后遍历弹出序列:
 1)如果弹出序列的当前元素temp比当前最大元素cur小，则它必须等于此时的栈顶元素top，否则要想得到和temp一样的值，就得不停弹出top，直到top == temp，但显然这样已经和弹出序列不匹配了，不可行，isSuccess置为false。
 2)如果temp比最大元素cur更大，则把cur + 1和temp之间的元素压入栈中，并检查此时栈容量是否超过要求，超过表明该序列也不可行,isSuccess置为false。

3.遍历完弹出序列后，根据标识isSuccess的值输出YES or No即可。

代码

#include<iostream>
#include<vector>
using namespace std;

int main()
{
     int N,M,K;
     while(cin >> M >> N >> K)
     {
         vector<int> stk;
         while(K--)
         {
           bool isSuccess = true;
           int curmax = 1;
           stk.push_back(1);
           for(int i = 1; i <= N;i++)
           {
               int temp;
               cin >> temp;
               if(temp > curmax)
               {
                   for(int j = curmax + 1; j <= temp;j++)
                    stk.push_back(j);
                   if(stk.size() > M)
                   {
                       isSuccess = false;
                   }
                   curmax = temp;
               }
               else
               {
                   if(stk.back() != temp)
                 {
                   isSuccess = false;
                 }
               }
               stk.pop_back();
           }
            if(isSuccess)
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
         }
     }
}