2020 C ++ priority_queue maximum and minimum heap heap introduction

1. Description of the problem

Usually in the brush theme, will encounter the largest heap, the minimum heap of problems, this time if they go to a realization is OK, but is usually not enough time, so how to deal with? At this point, you can make use of priority_queue C ++ STL.

2, concrete analysis

• It should be noted, C ++ STL default priority_queue is the biggest priority on the head of the queue, that is the biggest heap. So how to achieve the minimum heap it?

Suppose as a struct:

struct Node {
    int value;
    int idx;
    Node (int v, int i): value(v), idx(i) {}
    friend bool operator < (const struct Node &n1, const struct Node &n2) ; 
};

inline bool operator < (const struct Node &n1, const struct Node &n2) {
    return n1.value < n2.value;
}

priority_queue<Node> pq; // 此时pq为最大堆

If the minimum required heap, you need to achieve the following:

struct Node {
    int value;
    int idx;
    Node (int v, int i): value(v), idx(i) {}
//  friend bool operator < (const struct Node &n1, const struct Node &n2) ;
    friend bool operator > (const struct Node &n1, const struct Node &n2) ;
}; 

inline bool operator > (const struct Node &n1, const struct Node &n2) {
    return n1.value > n2.value;
}

priority_queue<Node, vector<Node>, greater<Node> > pq; // 此时greater会调用 > 方法来确认Node的顺序，此时pq是最小堆

3, otherwise

Of course, some of the smaller more hack means of. To construct such a minimum stack of type int:

priority_queue<int> pq; // 
pq.push( -1 * v1) ; //
pq.push( -1 * v2) ; //
pq.push( -1 * v3) ; // 分别是插入v1, v2, v3变量的相反数，那么pq其实也就变相成为了最小堆，只是每次从pq取值后，要再次乘以-1即可

4, with the smallest stack to solve the problem

Codeforces Round #536 (Div. 2), problem: (D) Lunar New Year and a Wander

Recommended reading this blog --Codeforces Round # 536, problem (Div 2.): (D) Lunar New Year and a Wander minimum heap [+ bfs]

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