poj1785&zoj2243 Binary Search Heap Construction(笛卡尔树)
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,...,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 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)))
Source: University of Ulm Local Contest 2004
题目大意：给n个二元组，每个二元组由一个字符串和一个数字组成，中间一个/分隔。现在要求按字符串为关键字构造一颗二叉搜索树，同时按数字为关键字构造一个大堆。
题目分析：红果果的笛卡尔树。
百科名片
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 50005;
struct node
{
int dig;
char cha[10];
int pa,l,r;
}lcm[N];
int stack[N];
char buf[100];
int n;
int cmp(struct node a,struct node b)
{
return strcmp(a.cha,b.cha) < 0;
}
int build()
{
int i,top,j;
for(i = 0;i < n;i ++)
lcm[i].l = lcm[i].r = lcm[i].pa = -1;
top = -1;
for(i = 0;i < n;i ++)
{
j = top;
while(j >= 0 && lcm[stack[j]].dig < lcm[i].dig)
j --;
if(j != -1)
{
lcm[i].pa = stack[j];
lcm[stack[j]].r = i;
}
if(j < top)
{
lcm[stack[j + 1]].pa = i;
lcm[i].l = stack[j + 1];
}
stack[++ j] = i;
top = j;
}
lcm[stack[0]].pa = -1;
return stack[0];
}
void dfs(int cur)
{
if(cur == -1)
return;
printf("(");
dfs(lcm[cur].l);
printf("%s/%d",lcm[cur].cha,lcm[cur].dig);
dfs(lcm[cur].r);
printf(")");
}
int main()
{
int i;
while(scanf("%d",&n),n)
{
for(i = 0;i < n;i ++)
{
scanf("%s",buf);
sscanf(buf,"%[^/]/%d",lcm[i].cha,&lcm[i].dig);
}
sort(lcm,lcm + n,cmp);
int root = build();
dfs(root);
putchar(10);
}
return 0;
}
//280ms 1748kb