【poj2828】Buy Tickets

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 next N lines contain the pairs of values Pos_i and Val_iin 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.

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

4
0 77
1 51
1 33
2 69
4
0 20523
1 19243
1 3890
0 31492

Sample Output

77 33 69 51
31492 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.

Source

POJ Monthly--2006.05.28, Zhu, Zeyuan

☆题目大意：给出每个人的位置，让你求出最后的排列顺序，插队情况，若位置相同则插队者在该位置，原位置上的人往后挪，就是插队的情形。

☆思路：若按着普通思路，会发现很麻烦的，那么如果逆向思维，最后插队的人的位置不会被别人影响，也就是确定的，在继续往前推，遇到重复位置的就往后放，因为逆向看的话，最晚出现在该位置的人一定在该位置。※树是储存位置数的，把序号0当作位置1，位置0与0个位置会出现问题。

☆样例：1.建树-（1,4）-（1,2）-（3,4）-（1）-（2）-（3）-（4）;

　　　　2.一般：3（2） 77，3>2（1,2）(在位置3前应有两个空位，若3放进去就只剩1个，所以放不进去，若有3个空位，则可将3放入第3个)，则向右子树遍历，※这时，位置3在去掉前两个位置后的相对位置为1（3-2）；

　　　　3.位置重合时：2（1）55，这时左子树只剩下1个空位，所以进不去。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #define MAXN 200000
 using namespace std;
 int segtree[MAXN*],a[MAXN],b[MAXN],c[MAXN];
 int n;
 void adddata(int now)
 {
     segtree[now]=segtree[(now<<)]+segtree[(now<<)+];
 }
 void buildtree(int now,int l,int r)
 {
     if (l==r)    {segtree[now]=;    return;}
     int mid=(l+r)>>;
     buildtree((now<<),l,mid);
     buildtree((now<<)+,mid+,r);
     adddata(now);
 }
 int query(int now,int l,int r,int v)
 {
     if (l==r)    {segtree[now]=;    return l;}
     int mid=(l+r)>>,ans;
     if (v<=segtree[(now<<)])    ans=query((now<<),l,mid,v);
     else    ans=query((now<<)+,mid+,r,v-segtree[(now<<)]);
     adddata(now);
     return ans;
 }
 int main()
 {
     int i,pos;
     while (~scanf("%d",&n))
     {
         for (i=;i<=n;i++)    scanf("%d%d",&a[i],&b[i]);
         buildtree(,,n);
         for (i=n;i>=;i--)
         {
             pos=query(,,n,a[i]+);
             c[pos]=b[i];
         }
         for (i=;i<n;i++)    printf("%d ",c[i]);
         printf("%d\n",c[n]);
     }
     return ;
 }