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() { 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 +  : ;
     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 = ;
 typedef pair<int, string> PIS;
 Leftist<PIS> pq[N];
 char buf[];

 int main() {
     int n, m;
     while (~scanf("%d%d", &n, &m)) {
         int k, x, y;
         for (int i = ; i <= n; i++) {
             pq[i].clear();
             scanf("%d", &k);
             for (int j = ; j < k; j++) {
                 scanf("%s%d", buf, &x);
                 pq[i].push(PIS(-x, buf));
             }
         }
         for (int i = ; 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 ;
 }

——written by Lyon