Buy Tickets（线段树单点更新，逆向思维）

Posted 2020-10-15 余生漫漫浪

题目大意：有n个的排队，每一个人都有一个val来对应，每一个后来人都会插入当前队伍的某一个位置pos。要求把队伍最后的状态输出。

个人心得：哈哈，用链表写了下，果不其然超时了，后面转念一想要用静态数组思维， 还是炸了。大牛们很给力，逆向一转，真是服气。

一想是呀，转过来的话那么此时的人必然可以得到他的位置，此时更新长度，后面的人在此时的队列中依旧可以得到他想要的位置。

就算思路知道了，怎么实现呢，大神果然不愧是大神，线段树sum表示总长度，节点表示是否存在被占据，然后更新就可以了。

真的是佩服到一派涂地，orz.....orz

题目原文：

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_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.

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.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<iomanip>
 6 #include<algorithm>
 7 using namespace std;
 8 #define inf 1<<29
 9 #define nu 4000005
10 #define maxnum 200005
11 int n;
12 int ans[maxnum],sum[maxnum<<2];
13 int pre[maxnum],val[maxnum];
14 void upset(int root)
15 {
16     sum[root]=sum[root*2]+sum[root*2+1];
17 }
18 void build(int root,int l,int r)
19 {
20     if(l==r)
21     {
22         sum[root]=1;
23         return ;
24     }
25     int mid=(l+r)/2;
26     build(root*2,l,mid);
27     build(root*2+1,mid+1,r);
28     upset(root);
29 }
30 void update(int root,int l,int r,int j)
31 {
32     if(l==r){
33         sum[root]--;
34         ans[l]=val[j];
35         return ;
36     }
37     int mid=(l+r)/2;
38     if(sum[root*2]>=pre[j])
39         update(root*2,l,mid,j);
40     else
41     {
42        pre[j]-=sum[root*2];
43        update(root*2+1,mid+1,r,j);
44     }
45     upset(root);
46 }
47 int main()
48 {
49     int i,j;
50     while(scanf("%d",&n)!=EOF){
51     for(i=1;i<=n;i++)
52         {
53            scanf("%d%d",&pre[i],&val[i]);
54             pre[i]++;
55         }
56     build(1,1,n);
57     for(i=n;i>0;i--)
58     {
59         update(1,1,n,i);
60     }
61    printf("%d",ans[1]);
62     for(i=2;i<=n;i++)
63        printf(" %d",ans[i]);
64        printf("\n");
65     }
66     return 0;
67 }