POJ 1785 Binary Search Heap Construction

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. 

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.

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.

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

(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)))

Ulm Local 2004

  看到这题后学的treap，没想到tle，了解到因为事先分配了优先级，复杂度可能退化到O(n*n),然后又学的笛卡尔树，区别在于笛卡尔树（只关注右链）优先级已经事先分配好、

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#define N 50010
using namespace std;
struct num
{
    char label[20];
    int val;
}a[N];
struct Num
{
    int ch[2];
    int val;
    char label[20];
}b[N];
char s1[20];
int n,stack[N],op[N];
int cmp(const void *p1,const void *p2)
{
    struct num * pt1 = (struct num *)p1;
    struct num * pt2 = (struct num *)p2;
    if(strcmp(pt1->label,pt2->label)>0)
    {
        return 1;
    }else
    {
        return -1;
    }
}
int main()
{
    //freopen("data.txt","r",stdin);
    int insert();
    void find(int k);
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        {
            break;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s1);
            int Top=0;
            int l = strlen(s1);
            int pos;
            for(int j=0;j<=l-1;j++)
            {
                if(s1[j]==‘/‘)
                {
                    pos = j;
                    a[i].label[Top]=‘\0‘;
                    break;
                }
                a[i].label[Top++] = s1[j];
            }
            int sum = 0;
            for(int j=pos+1;j<=l-1;j++)
            {
                sum = sum*10+s1[j]-‘0‘;
            }
            a[i].val = sum;
        }
        qsort(a+1,n,sizeof(a[0]),cmp);
        for(int i=1;i<=n;i++)
        {
            b[i].ch[0] = b[i].ch[1] = 0;
            b[i].val = a[i].val;
            strcpy(b[i].label,a[i].label);
        }
        int head= insert();
        find(head);
        printf("\n");
    }
    return 0;
}
int insert()
{
    int Top=-1;
    for(int i=1;i<=n;i++)
    {
        int k = Top;
        while(k!=-1&&a[stack[k]].val<a[i].val)
        {
            k--;
        }
        if(k!=-1)
        {
            op[i] = stack[k];
            b[stack[k]].ch[1] = i;
        }
        if(k!=Top)
        {
            op[stack[k+1]] = i;
            b[i].ch[0] = stack[k+1];
        }
        Top = k;
        stack[++Top] = i;
    }
    return stack[0];
}
void find(int k)
{
    if(k==0)
    {
        return ;
    }
    printf("(");
    find(b[k].ch[0]);
    printf("%s/%d",b[k].label,b[k].val);
    find(b[k].ch[1]);
    printf(")");
}