POJ-1785-Binary Search Heap Construction(笛卡尔树)

Posted 2020-09-21 yfceshi

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

思路：先按字符串排下序，建笛卡尔树，递归输出。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

struct S{
int id,val;
int parent,l,r;
}node[50005];

int stk[50005],top;
char s[50005][20];

bool cmp(struct S a,struct S b)
{
    if(strcmp(s[a.id],s[b.id])<0) return 1;

    return 0;
}

int build(int n)
{
    int i;

    top=0;

    stk[top]=0;

    for(i=1;i<n;i++)
    {
        while(top>=0 && node[stk[top]].val<node[i].val) top--;//注意。这里让小于当前值的出栈

        if(top>-1)
        {
            node[i].parent=stk[top];
            node[node[stk[top]].r].parent=i;
            node[i].l=node[stk[top]].r;
            node[stk[top]].r=i;
        }
        else
        {
            node[stk[0]].parent=i;
            node[i].l=stk[0];
        }

        stk[++top]=i;
    }

    return stk[0];//返回根节点
}

void dfs(int x)
{
    printf("(");//先输出左括号

    if(node[x].l!=-1) dfs(node[x].l);//假设有左子树。就先输出左子树

    printf("%s/%d",s[node[x].id],node[x].val);//输出自己

    if(node[x].r!=-1) dfs(node[x].r);//再输出右子树

    printf(")");//右括号
}

int main()
{
    int n,i,root;

    while(~scanf("%d",&n) && n)
    {
        for(i=0;i<n;i++)
        {
            scanf("%*[ ]%[a-z]/%d",s[i],&node[i].val);

            node[i].id=i;

            node[i].parent=node[i].l=node[i].r=-1;
        }

        sort(node,node+n,cmp);//按字符串字典序排序

        root=build(n);//建树

        dfs(root);//递归输出

        printf("\n");
    }
}