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;
         }
     }
}

 

 
posted @ 2017-10-03 01:36  0kk470  阅读(215)  评论(0编辑  收藏  举报