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).
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 Input
1
4
1 1
1 2
2 1
2 2
Sample Output
2
1 4
题意：有n个人，每一个人有一个Si和Bi，假设Si < Sj && Bi < Bj。则 i 和 j 不互相讨厌。问从这n个人中最多能够选出多少个人使得随意两个人都不会互相讨厌。
分析：由于一个人的S和B都严格小于另外一个人的S和B时，这连个人才不会互相讨厌。所以能够先对S从小到大排序，假设S同样则把B从大到小排序，这样问题就转化为了求最长上升子序列。当S同样时之所以要对B从大到小排序，是由于假设S递增的话，那么选出的方案中会将S同样的几个人都包括进去。不符合题目要求。然后就用O(nlogn)的算法求最长上升子序列的同一时候。记录到达每个数时的最大长度。最后从最大长度向下输出路径就可以。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100005;
int dp[N]; //dp[i]表示最长上升子序列长度为i时子序列末尾的最小B值
int mark[N]; //mark[i]表示以第i个people结尾的最长上升子序列的长度
struct Member {
int s, b;
int id;
bool operator < (const Member &x) const {
if(s == x.s) return x.b < b;
return s < x.s;
}
} a[N];
int main() {
int n;
while(~scanf("%d", &n)) {
for(int i = 1; i <= n; i++) {
scanf("%d%d", &a[i].s, &a[i].b);
a[i].id = i;
}
sort(a+1, a+n+1);
int max_len = 0;
dp[++max_len] = a[1].b;
mark[1] = max_len;
for(int i = 2; i <= n; i++) {
if(dp[max_len] < a[i].b) {
dp[++max_len] = a[i].b;
mark[i] = max_len;
}
else {
int k = lower_bound(dp+1, dp+1+max_len, a[i].b) - dp;
dp[k] = a[i].b;
mark[i] = k;
}
}
printf("%d
", max_len);
for(int i = n; i >= 1; i--) {
if(mark[i] == max_len) {
printf("%d", a[i].id);
if(max_len > 1) printf(" ");
max_len--;
}
}
printf("
");
}
return 0;
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