sgu 199 Beautiful People （DP-LIS O(nlogn)算法 输出所选元素）

199. Beautiful People

time limit per test: 0.25 sec.

memory limit per test: 65536 KB

input: standard

output: standard

The most prestigious sports club in one city has exactly N members. Each of its members is strong and beautiful. More precisely, i-th member of this club (members being numbered by the time they entered the club) has strength Si and beauty Bi .
Since this is a very prestigious club, its members are very rich and therefore extraordinary people, so they often extremely hate each other. Strictly speaking, i-th member of the club Mr X hates j-th member of the club Mr Y if Si ≤ Sj and
Bi ≥ Bj or if Si ≥ Sj and Bi ≤ Bj (if both properties of Mr X are greater then corresponding properties of Mr Y, he doesn't even notice him, on the other hand, if both of his properties are
less, he respects Mr Y very much).

To celebrate a new 2003 year, the administration of the club is planning to organize a party. However they are afraid that if two people who hate each other would simultaneouly attend the party, after a drink or two they would start a fight. So no two people
who hate each other should be invited. On the other hand, to keep the club presti≥ at the apropriate level, administration wants to invite as many people as possible.

Being the only one among administration who is not afraid of touching a computer, you are to write a program which would find out whom to invite to the party.

Input

The first line of the input file contains integer N — the number of members of the club. ( 2 ≤ N ≤ 100,000 ). Next N lines contain two numbers each — Si and Bi respectively ( 1 ≤ Si, Bi ≤ 109 ).

Output

On the first line of the output file print the maximum number of the people that can be invited to the party. On the second line output N integers — numbers of members to be invited in arbitrary order. If several solutions exist, output any one.

Sample test(s)

Input

4 1 1 1 2 2 1 2 2

Output

2

1 4

[align=right]大致题意：[/align]

从俱乐部中选一些人参加派对。每个人都有两个属性。S和B。一些属性高低难以分辨的人之间容易互相嫉妒，不适合一起参加派对。而派对想要邀请俱乐部中尽可能多的人，而每两人之间都需要满足s1<s2
和 b1<b2 才不会互相嫉妒。
[align=right]问最多能邀请到多少不会互相嫉妒的人，并且输出这些人的编号。[/align]

分析：
相当于选中的这些人的两个属性排起来都构成最长上升序列。
那么我们可以先按一个属性从小到大排序好，然后按另一个属性去找其最长上升序列。
因为数据比较大，用LIS的O（n^2）算法一定会超时，所以采用O（nlogn）算法来找LIS，并记录每个元素的前一个，方便输出。

LIS的O（nlogn）算法可以参考
http://www.slyar.com/blog/longest-ordered-subsequence.html~~~