标签：ecif gre cstring -o star length 容量 follow mes
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.
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.
For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.
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
YES
NO
NO
YES
NO
题意：
给定一个有固定容量的栈，1,2,...,n是入栈序列，元素出栈顺序随意，现给定出栈顺序(e.g.1~n的一个排列)，问这个出栈顺序是否合理，合理输出"YES"，否则输出"NO"。
题解：
开一个队列，开一个栈，输入一个数，就队列中这个数及之前的数放入栈中，放入不能超过容量。然后看栈顶元素是不是这个数，是就下一个，不是就标记NO。
AC代码：
#include<iostream> #include<stack> #include<queue> #include<cmath> #include<algorithm> #include<vector> #include<string> #include<cstring> using namespace std; int n,m,k; stack<int>s; queue<int>q; int main(){ cin>>n>>m>>k; while(k--){ while(!s.empty()) s.pop(); while(!q.empty()) q.pop(); int f=1; for(int i=1;i<=m;i++) q.push(i); for(int i=1;i<=m;i++){ int x; cin>>x; if(s.empty()||s.top()!=x){ while(!q.empty()){ if(s.size()<n) { //cout<<"把"<<q.front()<<"放栈"<<endl; s.push(q.front()); q.pop(); } else break; if(s.top()==x) break; } } if(s.top()==x){ s.pop(); //cout<<x<<"踢出栈"<<endl; continue; } f=0; } if(f) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }
PAT 甲级 1051 Pop Sequence (25 分)（模拟栈,较简单）
标签：ecif gre cstring -o star length 容量 follow mes
原文地址：https://www.cnblogs.com/caiyishuai/p/11483896.html
