Description
Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…
The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.
It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!
People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.
Input
There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The nextN lines contain the pairs of valuesPosi and Vali in the increasing order ofi (1 ≤i ≤ N). For each i, the ranges and meanings ofPosi andVali are as follows:
- Posi ∈ [0, i − 1] — The i-th person came to the queue and stood right behind thePosi-th person in the queue. The booking office was considered the 0th person and the person at the front of the queue was considered the first person in the queue.
- Vali ∈ [0, 32767] — The i-th person was assigned the valueVali.
There no blank lines between test cases. Proceed to the end of input.
Output
For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.
Sample Input
40 771 511 332 6940 205231 192431 38900 31492
Sample Output
77 33 69 5131492 20523 3890 19243
Hint
The figure below shows how the Little Cat found out the final order of people in the queue described in the first test case of the sample input.
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int n,ans[1000000];struct node{ int n,val;} a[1000000];struct tree{ int l,r,n;} b[1000000];void init(){ int i,j,k; for(k = 1; k<n; k<<=1); for(i = k; i<2*k; i++) { b[i].l = b[i].r = i-k+1; b[i].n = 1;//每个叶子节点只能放入一人 } for(i = k-1; i>0; i--) { b[i].l = b[2*i].l; b[i].r = b[2*i+1].r; b[i].n = b[2*i].n+b[2*i+1].n;//每个区间是其左右子树所能插入人数的总和 }}void insert(int i,int x,int m){ if(b[i].l == b[i].r)//找到叶子节点，这个节点存放该人，并且叶子节点能放的人数清0 { ans[b[i].l] = m; b[i].n = 0; return; } if(x<=b[2*i].n)//其插入的位置若能放入左子树（还能放人），往左子树方向搜，否则走右子树 insert(2*i,x,m); else insert(2*i+1,x-b[2*i].n,m); b[i].n--;//放完后，该节点的人数减一}int main(){ int i,j; while(~scanf("%d",&n)) { for(i = 1; i<=n; i++) scanf("%d%d",&a[i].n,&a[i].val); init(); for(i = n; i>0; i--)//逆推 { insert(1,a[i].n+1,a[i].val); } printf("%d",ans[1]); for(i = 2; i<=n; i++) printf(" %d",ans[i]); printf("\n"); } return 0;}