one way in，two way out 和 two way in， one way out

7-2 One Way In, Two Ways Out (25 分)
Consider a special queue which is a linear structure that allows insertions at one end, yet deletions at both ends. Your job is to check, for a given insertion sequence, if a deletion sequence is possible. For example, if we insert 1, 2, 3, 4, and 5 in order, then it is possible to obtain 1, 3, 2, 5, and 4 as an output, but impossible to obtain 5, 1, 3, 2, and 4.

Input Specification:
Each input file contains one test case. For each case, the first line gives 2 positive integers N and K (≤10), which are the number of insertions and the number of queries, respectively. Then N distinct numbers are given in the next line, as the insertion sequence. Finally K lines follow, each contains N inserted numbers as the deletion sequence to be checked.

All the numbers in a line are separated by spaces.

Output Specification:
For each deletion sequence, print in a line yes if it is indeed possible to be obtained, or no otherwise.

Sample Input:
5 4
10 2 3 4 5
10 3 2 5 4
5 10 3 2 4
2 3 10 4 5
3 5 10 4 2
Sample Output:
yes
no
yes
yes

题目意思：有个队列只能从一头插入，可以从两头删除。给你插入序列，让你判断删除序列是否正确。
思路：用一个双端队列模拟，依次插入，若队列两头有可删除的元素，则删除。最后检查一下删除序列有没有遍历完。

对于镜像问题two way in，one way out ，把它的删除序列当成插入序列，插入序列当成删除序列，当成one way in，two way out问题处理。

#include
#include //STL双端队列容器
#include
using namespace std;

bool judgmentSequence(vector<int> input, vector<int> output) {
    deque<int> seq;
    int index = 0;
    for (int i = 0; i < input.size(); i++) {
        // 限定双端队列尾部插入
        seq.push_back(input[i]);
        // 检查头尾能否删除
        while(seq.front() == output[index] || seq.back() == output[index]) {
            if (seq.front() == output[index]) {
                seq.pop_front();
                index++;
            }
            if (seq.back() == output[index]) {
                seq.pop_back();
                index++;
            }
        }
    }
    if (index == output.size())
        return true;
    return false;
}


int main() {
    vector<int> input = {10, 2, 3, 4, 5};
    vector<int> output = {3, 5, 10, 4, 2};
    if (judgmentSequence(input, output)) {
        cout << "yse" << endl;
    } else {
        cout << "no" << endl;
    }
    return 0;
}