hdu 1434 幸福列车 (Leftist Tree)

　　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：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <string>
 6 
 7 using namespace std;
 8 
 9 template<class T>
10 struct Node {
11     int d;
12     T dt;
13     Node *l, *r;
14     Node() { l = r = NULL;}
15     Node(T dt) : dt(dt), d(0) { l = r = NULL;}
16 } ;
17 
18 template<class T>
19 Node<T> *merge(Node<T> *a, Node<T> *b) {
20     if (!a) return b;
21     if (!b) return a;
22     if (a->dt < b->dt) return merge(b, a);
23     a->r = merge(a->r, b);
24     a->d = a->r ? a->r->d + 1 : 0;
25     return a;
26 }
27 
28 template<class T>
29 Node<T> *popTop(Node<T> *x) { Node<T> *ret = merge(x->l, x->r); delete x; return ret;}
30 
31 template<class T>
32 struct Leftist {
33     Node<T> *rt;
34     void clear() { rt = NULL;}
35     T top() { return rt->dt;}
36     void push(T dt) { rt = merge(rt, new Node<T>(dt));}
37     void pop() { rt = popTop(rt);}
38 } ;
39 
40 const int N = 11111;
41 typedef pair<int, string> PIS;
42 Leftist<PIS> pq[N];
43 char buf[111];
44 
45 int main() {
46     int n, m;
47     while (~scanf("%d%d", &n, &m)) {
48         int k, x, y;
49         for (int i = 1; i <= n; i++) {
50             pq[i].clear();
51             scanf("%d", &k);
52             for (int j = 0; j < k; j++) {
53                 scanf("%s%d", buf, &x);
54                 pq[i].push(PIS(-x, buf));
55             }
56         }
57         for (int i = 0; i < m; i++) {
58             scanf("%s%d", buf, &x);
59             if (!strcmp(buf, "GETOUT")) {
60                 puts(pq[x].top().second.c_str());
61                 pq[x].pop();
62             }
63             if (!strcmp(buf, "JOIN")) {
64                 scanf("%d", &y);
65                 pq[x].rt = merge(pq[x].rt, pq[y].rt);
66                 pq[y].rt = NULL;
67             }
68             if (!strcmp(buf, "GETON")) {
69                 scanf("%s%d", buf, &y);
70                 pq[x].push(PIS(-y, buf));
71             }
72         }
73     }
74     return 0;
75 }

View Code

——written by Lyon