ZOJ 2243 Binary Search Heap Construction笛卡尔树（二叉搜索+堆）



Binary Search Heap Construction

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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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_ndenoting 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/17 a/1 b/2 c/3 d/4 e/5 f/6 g/77 a/3 b/6 c/4 d/7 e/2 f/5 g/10

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：每一个节点有两个属性，“标号-优先级”，相对于标号是二叉搜索树，相对与优先级是大根堆 然后中根遍历

分析： 对标号进行从小到大排序，此时已是中根遍历了，然后在构造堆，插入的节点如果比他的父节点优先级小就放在右儿子位置。如果大的话就往上找直到找打比他大的那个节点记作x，此时x原来的右儿子更新为该节点的左儿子（中序遍历左儿子在前），该节点成为x的新右儿子。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int INF = 100000000;struct node{    int lson,rson,fa;    int v;    char s[100];};node tree[50010];int cmp(node x,node y){    return strcmp(x.s, y.s) < 0;    //字符串比较用这个；换成return x.s < y.s 就WA了}void inset_node(int now){    int j = now - 1;    while(tree[j].v < tree[now].v)        j = tree[j].fa;    tree[now].lson = tree[j].rson;    tree[j].rson = now;    tree[now].fa = j;}void travel(int now){    if(now == 0) return ;    printf("(");    travel(tree[now].lson);    printf("%s/%d",tree[now].s,tree[now].v);    travel(tree[now].rson);    printf(")");}int main(){    int n;    while(scanf("%d", &n) != EOF && n)    {        getchar();        for(int i = 1; i <= n; i++)        {            scanf("%[a-z]/%d",tree[i].s,&tree[i].v);            getchar();            tree[i].lson = tree[i].rson = tree[i].fa = 0;        }        sort(tree + 1,tree + n + 1, cmp);        tree[0].fa = tree[0].lson = tree[0].rson = 0;        tree[0].lson = 0;        tree[0].v = INF;        for(int i = 1; i <= n; i++)            inset_node(i);        travel(tree[0].rson);        printf("\n");    }    return 0;}