相应POJ 题目:点击打开链接
Binary Search Heap Construction
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 9075 | Accepted: 2566 |
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.
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)))
题意:
每次有n个输入,每一个输入格式为(字符串/数字)。字符串(长度未知,反正我开100也能过)和数字都不会反复。要求建立一棵树。使得中序遍历按字符串字典序排序,并且数字符合大根堆。
输出格式为((左子树)根节点(右子树))。
思路:
赤裸裸的Treap树,可惜会TLE。可用笛卡尔树顺利AC。建树时在右链从下往上找适合位置插入。读入的时候有点技巧,%*[ ]表示忽略[]里面的字符,%[^/]表示读入字符串时遇到'/'就结束,没有读入'/'且会在字符串后面加入结束符。
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <algorithm> #include <time.h> using namespace std; #define N 50010 #define inf 0x7fffffff #define nil 0 struct Node { int pri, l, r, fa; char str[100]; }; bool cmp(Node n1, Node n2) { return strcmp(n1.str, n2.str) < 0; } class CartesianTree { public: void Init(int n) { a[0].pri = inf; a[0].l = a[0].r = a[0].fa = nil; int i; for(i = 1; i <= n; i++){ scanf("%*[ ]%[^/]/%d", a[i].str, &a[i].pri); a[i].l = a[i].r = a[i].fa = nil; } sort(a + 1, a + n + 1, cmp); for(i = 1; i <= n; i++) Insert(i); } void Insert(int p) { int t = p - 1; //从下往上找 while(a[t].pri < a[p].pri) t = a[t].fa; a[p].l = a[t].r; a[t].r = p; a[p].fa = t; } void Show() { InOrder(a[0].r); printf("\n"); } void InOrder(int t) { if(nil == t) return; printf("("); InOrder(a[t].l); printf("%s/%d", a[t].str, a[t].pri); InOrder(a[t].r); printf(")"); } private: Node a[N]; }; CartesianTree ct; int main() { //freopen("in.txt","r",stdin); int n; while(scanf("%d", &n), n) { ct.Init(n); ct.Show(); } return 0; }