PAT-2019年冬季考试-甲级 7-4 Cartesian Tree

7-4 Cartesian Tree (30分)

A Cartesian tree is a binary tree constructed from a sequence of distinct numbers. The tree is heap-ordered, and an inorder traversal returns the original sequence. For example, given the sequence { 8, 15, 3, 4, 1, 5, 12, 10, 18, 6 }, the min-heap Cartesian tree is shown by the figure.

Your job is to output the level-order traversal sequence of the min-heap Cartesian tree.

Input Specification:

Each input file contains one test case. Each case starts from giving a positive integer N (≤30), and then N distinct numbers in the next line, separated by a space. All the numbers are in the range of int.

Output Specification:

For each test case, print in a line the level-order traversal sequence of the min-heap Cartesian tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the beginning or the end of the line.

Sample Input:

10
8 15 3 4 1 5 12 10 18 6

Sample Output:

1 3 5 8 4 6 15 10 12 18

Solution:

一看名字差点吓尿，这不是Treap的父类笛卡尔树吗？！

认真读题发现只要根据序列建树+遍历，这也太水了吧==

P.S. 其实也可以不用建树，在递归的时候给各节点打上层次标记即可~

#include <bits/stdc++.h>
using namespace std;

const int maxn = 50;
const int inf = 0x3f3f3f3f;
int inOrder[maxn];
struct node
{
    int val;
    node *ls, *rs;
    node(int v)
    {
        this->val = v;
        this->ls = this->rs = nullptr;
    }
};

inline const int read()
{
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = (x << 3) + (x << 1) + ch - '0'; ch = getchar(); }
    return x * f;
}

node* build(int l, int r)
{
    if (l > r) return nullptr;
    int rtid = 0, rtval = inf;
    for (int i = l; i <= r; i++)
    {
        if (inOrder[i] < rtval)
        {
            rtval = inOrder[i];
            rtid = i;
        }
    }
    node* now = new node(rtval);
    int nl = rtid - l + 1, nr = r - rtid + 1;
    if (nl > 0) now->ls = build(l, rtid - 1);
    if (nr > 0) now->rs = build(rtid + 1, r);
    return now;
}

void print(node* root)
{
    queue<node*> q;
    q.push(root);
    bool flag = false;
    while (!q.empty())
    {
        node* now = q.front();
        q.pop();
        printf("%s%d", flag ? " " : "", now->val);
        flag = true;
        if (now->ls) q.push(now->ls);
        if (now->rs) q.push(now->rs);
    }
}

int main()
{
    int n = read();
    for (int i = 1; i <= n; i++) inOrder[i] = read();
    node* root = build(1, n);
    print(root);
    return 0;
}