1051. 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 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

IDEA

1.用栈进行模拟

2.maxnum初始化为m，栈的容量，若遇到目标值大于当前maxnum，则是非法的

3.当栈为空，或者栈不空且栈顶元素不等于目标值且应该入栈的元素不大于n且栈中元素个数不大于栈容量时，num进栈

CODE

#include<iostream>#include<vector>#include<stack>#include<fstream>using namespace std;int main(){freopen("input.txt","r",stdin);int m,n,k;cin>>m>>n>>k;while(k--){int flag=1;vector<int> vec(n);for(int i=0;i<n;i++){cin>>vec[i];}stack<int> sta;int maxnum=m;int num=1;for(int i=0;i<n;i++){if(vec[i]>maxnum){cout<<"NO"<<endl;flag=0;break;}while ((sta.empty()||(!sta.empty()&&sta.top()!=vec[i]))&&num<=n&&sta.size()<=m){sta.push(num++);}if(sta.top()!=vec[i]){cout<<"NO"<<endl;flag=0;break;}sta.pop();maxnum++;}if(flag){cout<<"YES"<<endl;}}fclose(stdin);return 0;}