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 S _i and beauty B _i . 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 S _i ≤ S _j and B _i ≥ B _j or if S _i ≥ S _j and B _i ≤ B _j (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 — S _i and B _i respectively ( 1 ≤ S _i, B _i ≤ 10 ⁹ ). 

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 

大致题意：

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。

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。

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~~~

#include 
    
     
#include 
     
      
#include 
      
       
using namespace std;

int fa[100010],stk[100010],ret;

struct node{
	int a,b,id;
}men[100010];//记录每个人的s（a）和b信息，并且记录下是第几个人。

bool cmp(node x,node y)//排序先按a升序，若相同 再按b排序。
{
	if(x.a==y.a)
	{
		return x.b>y.b;
	}
	else
		return x.a
       
        men[stk[ret]].b)//比前一个元素大的数就将其选进序列
			{
				stk[++ret]=i;
				fa[i]=stk[ret-1];//为记录lis中这个元素的前一个是哪个，为了最后输出选择了哪些人。
			}
			else//二分检索现选择的最长上升子列中比这个元素大的第一个数，并将其替换。
			{
				int begin,end;
				begin=1;
				end=ret;
				while(begin<=end)
				{
					int mid=(begin+end)/2;
					if(men[i].b>men[stk[mid]].b)
					{
						begin=mid+1;
					}
					else
						end=mid-1;
				}
				stk[begin]=i;//替换，并记录其前一个人。
				fa[i]=stk[begin-1];
			}
		}
		printf("%d\n",ret);
		print(stk[ret]);//递归输出选中参加party的人。
	}
	return 0;
}