zoj 2243 Binary Search Heap Construction（笛卡尔树）

最新推荐文章于 2018-10-09 20:29:52 发布

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#笛卡尔树

本文详细介绍了如何使用特定的排序和构造方法来构建一棵笛卡尔树，其中包括了节点的优先级排序、构建过程以及输出格式的具体实现。

Binary Search Heap Construction

Time Limit: 5 Seconds Memory Limit: 32768 KB

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 Specification

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,...,l_n/p_n 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 Specification

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)))

题意：给出n个结点，有两个值key和value，要求构造一棵笛卡尔树。光看key的话，笛卡尔树是一棵二叉搜索树，每个节点的左子树的key都比它小，右子树都比它大；光看value的话，笛卡尔树有点类似堆，根节点的value是最小（或者最大）的，每个节点的value都比它的子树要小（或者大）。

思路：先按key值排序，然后一个个插入构造笛卡尔树，这里用了O（n）的算法求出了每个结点的父亲结点，然后对于每个结点，若该节点key值比父亲结点小，则为左子树，否则为右子树。

AC代码：

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

const int maxn = 50005;

int n;
struct node{
    char s[105];
    int v;
}p[maxn];
int l[maxn], r[maxn], T[maxn], st[maxn];
bool cmp(node a, node b){
    return strcmp(a.s, b.s) < 0;
}
void solve(){
    int k, top = -1;
    memset(l, -1, sizeof(l));
    memset(r, -1, sizeof(r));
    for(int i = 0; i < n; i++)
    {
        k = top;
        while(k >= 0 && p[st[k]].v < p[i].v) k--;
        if(k != -1) T[i] = st[k];
        if(k < top) T[st[k + 1]] = i;
        st[++k] = i;
        top = k;
    }
    T[st[0]] = -1;
    for(int i = 0; i < n; i++)
    {
        if(T[i] == -1) continue;
        if(strcmp(p[i].s, p[T[i]].s) < 0) l[T[i]] = i;
        else r[T[i]] = i;
    }
}
void dfs(int u){
    if(u == -1) return;
    printf("(");
    dfs(l[u]);
    printf("%s/%d", p[u].s, p[u].v);
    dfs(r[u]);
    printf(")");
}
int main()
{
    while(scanf("%d", &n), n)
    {
        getchar();
        for(int i = 0; i < n; i++)
        {
            scanf("%[^/]s", p[i].s);
            scanf("/%d", &p[i].v);
            getchar();
        }
        sort(p, p + n, cmp);
        solve();
        dfs(st[0]);
        puts("");
    }
    return 0;
}