POJ2828Buy Tickets

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.

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 next N lines contain the pairs of values Pos_i and Val_i in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Pos_i and Val_i are as follows:

• Pos_i ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Pos_i-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.
• Val_i ∈ [0, 32767] — The i-th person was assigned the value Val_i.

There no blank lines between test cases. Proceed to the end of input.

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.

题意：给定n个人的有序入队位置，即第i+1行描述了第i个人插到第几个人后面，编号是多少。

思路： 托dada的福，很快就敲定了思路。又一次用到了倒着做的思想，先读入所有的数据，然后倒着插，因为这时候，描述每个人的第一个数字就是指这个人前面的空位数，而这个空位数就要用到线段树来维护了。tree表示的是这个区间内的空位数。每次插入的时候，就将这个点插到相对应的位置（若空格数小于左儿子的，就插入做儿子的区间；若大于左儿子，就将空格数减去左儿子的空格数再插入右儿子的区间），并更新tree。

code：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int tree[800001]={0},ans[200001]={0},a[200001][2]={0};
void build(int i,int l,int r)
{
 int mid;
 if (l==r)
 {
  tree[i]=1;
  return;
 }
 mid=(l+r)/2;
 build(i*2,l,mid);
 build(i*2+1,mid+1,r);
 tree[i]=tree[i*2]+tree[i*2+1];
}
void insert(int i,int l,int r,int aa,int k)
{
 int mid;
 if (l==r)
 {
  ans[l]=k;
  tree[i]=0;
  return;
 }
 mid=(l+r)/2;
 if (aa<=tree[i*2]) insert(i*2,l,mid,aa,k);
 else insert(i*2+1,mid+1,r,aa-tree[i*2],k);
 tree[i]=tree[i*2]+tree[i*2+1];
}
int main()
{
 int i,j,n;
 while(scanf("%d",&n)==1)
 {
  memset(ans,0,sizeof(ans));
  memset(tree,0,sizeof(tree));
     build(1,1,n);
  for (i=1;i<=n;++i)
       scanf("%d%d",&a[i][0],&a[i][1]);
     for (i=n;i>=1;--i)
       insert(1,1,n,a[i][0]+1,a[i][1]);
     for (i=1;i<n;++i)
       printf("%d ",ans[i]);
        printf("%d\n",ans[n]);
 }
}