02-线性结构4 Pop Sequence (25分)
Given a stack which can keep MM numbers at most. Push NN numbers in the order of 1, 2, 3, …, NN 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 MM is 5 and NN 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): MM (the maximum capacity of the stack), NN (the length of push sequence), and KK (the number of pop sequences to be checked). Then KK lines follow, each contains a pop sequence of NN 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
AC代码:
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
int M; //栈的最大容量
int N; //测试数据的长度
int K; //测试数据的组数
/*
这题是模拟给定序列出栈过程
思路如下：
当我们遇见输出x时，则要考虑的是x前的元素，即小于等于x的元素都先push栈，才会有pop x；
1。栈为空时，判断需要填入的数 是否小于 栈的容量(即M)
2。若后一个数比前一个数大，又要push其之前的数 再判断
3。若后一个数比前一个数小，则要判断栈顶元素是否与其相等
*/
bool Check_Seq(vector<int> &v)
{
int i = 0;
int num = 1;
int cap = M+1;
stack<int> sta;
sta.push(0);
while(i < N ){
while(sta.size() < cap && v[i] > sta.top()){
sta.push(num++);
}
if ( v[i++] == sta.top()){
sta.pop();
}else{
return false;
}
}
return true;
}
int main()
{
cin>>M>>N>>K;
vector<int> seq;
for (int i = 0 ; i < K ; i++){
for ( int j = 0 ; j < N ; j++){
int tmp;
cin>>tmp;
seq.push_back(tmp);
}
if(Check_Seq(seq)){
cout<<"YES"<<endl;
}else{
cout<<"NO"<<endl;
}
seq.clear();
}
return 0;
}腾讯云开发者
