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

Ulm Local 2004

 

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

 

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=100010;
struct node{
    char s[30];
    int val;
    int fa,l,r;
    bool operator < (node b)const{
        return strcmp(s,b.s)<0;
    }
}t[mxn];
void init(int n){
    for(int i=0;i<=n;i++){
        t[i].l=t[i].r=t[i].fa=0;
    }
    t[0].val=0x3f3f3f3f;
}
int n;
//int st[mxn],top=0;
void Build(){
    for(int i=1;i<=n;i++){
        int j=i-1;
        while(t[j].val<t[i].val)
            j=t[j].fa;
        t[i].l=t[j].r;t[j].r=i;
        t[i].fa=j;
    }
    return;
}
void Print(int u){
    if(!u)return;
    printf("(");
    Print(t[u].l);
    printf("%s/%d",t[u].s+1,t[u].val);
    Print(t[u].r);
    printf(")");
    return;
}
int main(){
    int i,j;
    while(scanf("%d",&n) && n){
        init(n);
        for(i=1;i<=n;i++)
            scanf("%[^/]/%d",t[i].s,&t[i].val);
        sort(t+1,t+n+1);
        Build();
        Print(t[0].r);
        printf("\n");
    }
    return 0;
}