• 时间限制：400ms
• 内存限制：64MB
• 代码长度限制：16kB
• 判题程序：系统默认
• 作者：陈越
• 单位：浙江大学

https://pta.patest.cn/pta/test/3512/exam/4/question/62615

Given a stack which can keep MMM numbers at most. Push NNN numbers in the order of 1, 2, 3, ..., NNN 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 MMM is 5 and NNN 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): MMM (the maximum capacity of the stack), NNN (the length of push sequence), and KKK (the number of pop sequences to be checked). Then KKK lines follow, each contains a pop sequence of NNN 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

用回溯法（dfs）判断能否构造出目标序列

 #include<bits/stdc++.h>

 using namespace std;

 int gode[];

 int M,N,K,num;

 stack<int> Sta;

 bool dfs(int ipush,int ipop)//ipush,ipop为已完成的操作(push,pop)数

 {

     if(ipop==N)//成功构造gode

         return true;

     if(ipush<N&&Sta.size()<M)

     {

         Sta.push(num++);

         if(dfs(ipush+,ipop)) return true;

         num--;

         Sta.pop();

     }

     if(ipop<ipush&&!Sta.empty()&&Sta.top()==gode[ipop])//栈顶元素即为gode的下一个数时才执行pop

     {

         int t=Sta.top();

         Sta.pop();

         if(dfs(ipush,ipop+)) return true;

         Sta.push(t);

     }

     return false;

 }

 int main()

 {

     cin>>M>>N>>K;

     for(int i=;i<K;i++)

     {

         num=;

         for(int j=;j<N;j++)

             cin>>gode[j];

         puts(dfs(,)?"YES":"NO");

     }

     return ;

 }