1.3.46

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question;

Forbidden triple for stack generability. Prove that a permutation can be generated by a stack (as in the previous question ) if and only if it has no forbidden triple (a,b,c) such that a < b < c with c first, a second, and b third (possibly with other intervening integers between c and a and between a and b).

answer:

//感觉就像是对上一题中generated()的解释

Partial solution: Suppose that there is a forbidden triple(a,b,c). Item c is popped before a and b, but a and b are pushed before c. Thus, when c is pushed, both a and b are on the stack. Therefore, a cannot be popped before b.

1.3.46

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原文地址：https://www.cnblogs.com/w-j-c/p/9093050.html