PAT 数据结构 02-线性结构4. Pop Sequence (25)

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 51 2 3 4 5 6 73 2 1 7 5 6 47 6 5 4 3 2 15 6 4 3 7 2 11 7 6 5 4 3 2

Sample Output:

YESNONOYESNO

若穷举出所有可能序列，再一一进行比较，效率太低。

直接对目标序列进行检测，模拟操作。

/*2015.7.8*/#include <iostream>#include <vector>#include <stack>#include <string>#include <sstream>#include <iomanip>#include <math.h>using namespace std;// pop seuenceint check(const vector<int> &ivec,int M,int N){stack<int> stk;int i=0;int j=1;while(1){while(stk.empty()||stk.top()!=ivec[i]){stk.push(j++);if(stk.size()==M){  if(stk.top()==ivec[i]){  stk.pop();  i++;  if(i==N){  if(stk.empty())  cout<<"YES"<<endl;  else  cout<<"NO"<<endl;  return 0;  }  }else{  cout<<"NO"<<endl;  return 0;  }}}if(!stk.empty()){stk.pop();i++;}if(i==N&&stk.empty()){cout<<"YES"<<endl;return 0;}}return 0;}int main(){int M,N,K;cin>>M>>N>>K;vector<int> ivec(N);for(int i=0;i<K;i++){for(int i=0;i<N;i++)cin>>ivec[i];check(ivec,M,N);}return 0;}