hdu 1434 幸福列车 (Leftist Tree)

Problem - 1434

　　网上题解是普通的堆合并，都是用优先队列直接做的。可是正解的堆合并应该是用左偏堆或者斐波那契堆的吧，不然O(X * N ^ 2)的复杂度应该是过不了的。斐波那契堆的实现相对麻烦，所以我用了左偏堆完成这题，最坏复杂度O(X * N log N)。

　　这题就是一个模拟，没什么可以解释的。1y~

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

template<class T>
struct Node {
    int d;
    T dt;
    Node *l, *r;
    Node() { l = r = NULL;}
    Node(T dt) : dt(dt), d(0) { l = r = NULL;}
} ;

template<class T>
Node<T> *merge(Node<T> *a, Node<T> *b) {
    if (!a) return b;
    if (!b) return a;
    if (a->dt < b->dt) return merge(b, a);
    a->r = merge(a->r, b);
    a->d = a->r ? a->r->d + 1 : 0;
    return a;
}

template<class T>
Node<T> *popTop(Node<T> *x) { Node<T> *ret = merge(x->l, x->r); delete x; return ret;}

template<class T>
struct Leftist {
    Node<T> *rt;
    void clear() { rt = NULL;}
    T top() { return rt->dt;}
    void push(T dt) { rt = merge(rt, new Node<T>(dt));}
    void pop() { rt = popTop(rt);}
} ;

const int N = 11111;
typedef pair<int, string> PIS;
Leftist<PIS> pq[N];
char buf[111];

int main() {
    int n, m;
    while (~scanf("%d%d", &n, &m)) {
        int k, x, y;
        for (int i = 1; i <= n; i++) {
            pq[i].clear();
            scanf("%d", &k);
            for (int j = 0; j < k; j++) {
                scanf("%s%d", buf, &x);
                pq[i].push(PIS(-x, buf));
            }
        }
        for (int i = 0; i < m; i++) {
            scanf("%s%d", buf, &x);
            if (!strcmp(buf, "GETOUT")) {
                puts(pq[x].top().second.c_str());
                pq[x].pop();
            }
            if (!strcmp(buf, "JOIN")) {
                scanf("%d", &y);
                pq[x].rt = merge(pq[x].rt, pq[y].rt);
                pq[y].rt = NULL;
            }
            if (!strcmp(buf, "GETON")) {
                scanf("%s%d", buf, &y);
                pq[x].push(PIS(-y, buf));
            }
        }
    }
    return 0;
}

——written by Lyon