POJ1785 Binary Search Heap Construction

Time Limit: 2000MS   Memory Limit: 30000K
Total Submissions: 10309   Accepted: 2838

Description

Read the statement of problem G for the definitions concerning trees. In the following we define the basic terminology of heaps. A heap is a tree whose internal nodes have each assigned a priority (a number) such that the priority of each internal node is less than the priority of its parent. As a consequence, the root has the greatest priority in the tree, which is one of the reasons why heaps can be used for the implementation of priority queues and for sorting. 

A binary tree in which each internal node has both a label and a priority, and which is both a binary search tree with respect to the labels and a heap with respect to the priorities, is called a treap. Your task is, given a set of label-priority-pairs, with unique labels and unique priorities, to construct a treap containing this data. 

Input

The input contains several test cases. Every test case starts with an integer n. You may assume that 1<=n<=50000. Then follow n pairs of strings and numbers l1/p1,...,ln/pn denoting the label and priority of each node. The strings are non-empty and composed of lower-case letters, and the numbers are non-negative integers. The last test case is followed by a zero.

Output

For each test case output on a single line a treap that contains the specified nodes. A treap is printed as (< left sub-treap >< label >/< priority >< right sub-treap >). The sub-treaps are printed recursively, and omitted if leafs.

Sample Input

7 a/7 b/6 c/5 d/4 e/3 f/2 g/1
7 a/1 b/2 c/3 d/4 e/5 f/6 g/7
7 a/3 b/6 c/4 d/7 e/2 f/5 g/1
0

Sample Output

(a/7(b/6(c/5(d/4(e/3(f/2(g/1)))))))
(((((((a/1)b/2)c/3)d/4)e/5)f/6)g/7)
(((a/3)b/6(c/4))d/7((e/2)f/5(g/1)))

Source

树 笛卡尔树模板题

笛卡尔树满足两个性质:中序遍历就是原数组(二叉搜索树性质),以及数值满足堆的性质(上小下大或上大下小)。

基本上就是没加随机旋转的treap

这题把读入数据建成笛卡尔树即可,要求字符串满足二叉树性质,数值满足大根堆性质

strcmp跑得真够慢

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 using namespace std;
 9 const int mxn=100010;
10 struct node{
11     char s[30];
12     int val;
13     int fa,l,r;
14     bool operator < (node b)const{
15         return strcmp(s,b.s)<0;
16     }
17 }t[mxn];
18 void init(int n){
19     for(int i=0;i<=n;i++){
20         t[i].l=t[i].r=t[i].fa=0;
21     }
22     t[0].val=0x3f3f3f3f;
23 }
24 int n;
25 //int st[mxn],top=0;
26 void Build(){
27     for(int i=1;i<=n;i++){
28         int j=i-1;
29         while(t[j].val<t[i].val)
30             j=t[j].fa;
31         t[i].l=t[j].r;t[j].r=i;
32         t[i].fa=j;
33     }
34     return;
35 }
36 void Print(int u){
37     if(!u)return;
38     printf("(");
39     Print(t[u].l);
40     printf("%s/%d",t[u].s+1,t[u].val);
41     Print(t[u].r);
42     printf(")");
43     return;
44 }
45 int main(){
46     int i,j;
47     while(scanf("%d",&n) && n){
48         init(n);
49         for(i=1;i<=n;i++)
50             scanf("%[^/]/%d",t[i].s,&t[i].val);
51         sort(t+1,t+n+1);
52         Build();
53         Print(t[0].r);
54         printf("
");
55     }
56     return 0;
57 }
原文地址:https://www.cnblogs.com/SilverNebula/p/6686584.html