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 2005 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 prestige 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.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
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).
<p< dd=""> 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.
<p< dd=""> Sample Input
1
4
1 1
1 2
2 1
2 2
<p< dd=""> Sample Output
2
1 4
题意：如果S1<=S2 && beaut1>=beaut2 或 S1>=S2 && beaut1<=beaut2那他们就是敌人，不可以一起出现在party，且要满足如果一号人的综合实力大于二号，且二号大于三号，那么一号和三号就不是敌人
思路：我们可以将条件转换一下，题上所给的条件比较复杂，但是转化为s1>s2 && beaut1>beaut2这样就感觉清晰了，这不就是最长上升子序列嘛
于是我们可以先对他的一个属性排序，当着一个属性相同时再对另一个属性反向排序（就是这两个属性的排序方式不同）
但是求最长上升子序列不能用平常的方法，要用优化的nlogn的方法（二分法）
但是二分法求出来的序列并不符合题意，但是它的长度是符合题意得，所以我们要在这个过程中记录其中间的变化，以来求出来符合题意的序列
要注意的是：二分法求出来的序列本来就不符合题意，如果有的例子过了，要就一定是碰巧，真正的序列要另开一个数据来存
举个例子：
输入：
1 2
1 3
3 12
2 9
2 7
3 10
2 1
对这几个数据的s进行从小到大排序，beaut进行从大到小
排序后：
1 3
1 2
2 9
2 7
2 1
3 12
3 10
按照二分法我们要一个dp数组来存二分求出来的序列（不符合题意得序列），和一个记录标号的数组pre
此时只需要求beaut的最长上升子序列
dp 3
pre 1
dp 2 2小于三所以要替换他的位置，以达到上升子序列的目的
pre 1 1
dp 2 9
pre 1 1 2
dp 2 7
pre 1 1 2 2
dp 1 7 找的是刚好大于或等于此值的位置，并和其交换
pre 1 1 2 2 1
dp 1 7 12
pre 1 1 2 2 1 3
dp 1 7 10
（头）pre 1 1 2 2 1 3 3（尾）
此时最长上升子序列的长度是3（用k来代替），所以我们要从pre数组的n循环找到1，从中找到符合题意的序列
从pre数组尾开始，如果里面存的值等于k，就输出此处序列下标所对应的原输入的序号（看一下代码理解一下），再让k--
代码如下：
1 #include<stdio.h> 2 #include<string.h> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 const int INF=0xfffffff; 7 #define N 100000+10 8 #define inf 0x3f3f3f3f 9 struct shudui 10 { 11 int id; 12 int st,value; 13 }m[N]; 14 bool cmp(shudui n1,shudui n2) 15 { 16 if(n1.st==n2.st) 17 return n1.value>n2.value; //用小的b值来覆盖大的b值 18 return n1.st<n2.st; 19 } 20 int pre[100005],dp[100005]; 21 int main() 22 { 23 int t; 24 scanf("%d",&t); 25 while(t--) 26 { 27 int n; 28 scanf("%d",&n); 29 for(int i=1; i<=n; ++i) 30 { 31 scanf("%d%d",&m[i].st,&m[i].value); 32 m[i].id=i; 33 } 34 memset(dp,inf,sizeof(dp)); //不能用INF来初始化数据，如果用的话，数组里面的值全部是-1 35 // for(int i=1;i<=n;++i) 36 // printf("%d ",dp[i]); 37 // printf("\n"); 38 sort(m+1,m+1+n,cmp); 39 //int g1,g2,g3; 40 41 int k=0; 42 // for(int i=1;i<=n;++i) 43 // { 44 // int j=lower_bound(dp+1,dp+1+n,m[i].value)-dp; 45 // dp[j]=m[i].value; 46 // pre[i]=j; 47 // k=max(k,j); 48 // } 49 // printf("%d\n",k); 50 // for(int i=n;i>=1;--i) 51 // { 52 // if(pre[i]==k) 53 // { 54 // printf("%d ",m[i].id); 55 // k--; 56 // } 57 // } 58 // printf("\n"); 59 // if(t!=0) printf("\n"); 60 for(int i=1; i<=n; i++) 61 { 62 int j=lower_bound(dp+1,dp+1+n,m[i].value)-dp; 63 dp[j]=m[i].value; 64 pre[i]=j; 65 k=max(k,j); 66 } 67 printf("%d\n",k); 68 for(int i=n; i>=1; i--) 69 { 70 if (pre[i]==k) 71 { 72 printf("%d ",m[i].id); 73 k--; 74 } 75 } 76 printf("\n"); 77 if(t!=0) 78 { 79 printf("\n"); 80 } 81 } 82 return 0; 83 }