Codeforces 754 D Fedor and coupons

D. Fedor and coupons
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket.
The goods in the supermarket have unique integer ids. Also, for every integer there is a product with id equal to this integer. Fedor has ndiscount coupons, the i-th of them can be used with products with ids ranging from li to ri, inclusive. Today Fedor wants to take exactly kcoupons with him.
Fedor wants to choose the k coupons in such a way that the number of such products x that all coupons can be used with this product x is as large as possible (for better understanding, see examples). Fedor wants to save his time as well, so he asks you to choose coupons for him. Help Fedor!
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — the number of coupons Fedor has, and the number of coupons he wants to choose.
Each of the next n lines contains two integers li and ri ( - 109 ≤ li ≤ ri ≤ 109) — the description of the i-th coupon. The coupons can be equal.
Output
In the first line print single integer — the maximum number of products with which all the chosen coupons can be used. The products with which at least one coupon cannot be used shouldn't be counted.
In the second line print k distinct integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the ids of the coupons which Fedor should choose.
If there are multiple answers, print any of them.
Examples
input

4 2
1 100
40 70
120 130
125 180

output

31
1 2 

input

3 2
1 12
15 20
25 30

output

0
1 2 

input

5 2
1 10
5 15
14 50
30 70
99 100

output

21
3 4 

Note
In the first example if we take the first two coupons then all the products with ids in range [40, 70] can be bought with both coupons. There are 31 products in total.
In the second example, no product can be bought with two coupons, that is why the answer is 0. Fedor can choose any two coupons in this example.
제목:
n 개 구간 을 드 리 겠 습 니 다. 마침 k 개 구간 에 교차 하 는 최대 구간 을 찾 아 보 세 요.
문제 풀이:
최종 구간 은 틀림없이 어느 구간 의 왼쪽 단점 과 어느 구간 의 오른쪽 단점 으로 구성 되 어 있 기 때문에 우 리 는 구간 을 정렬 한 후 왼쪽 에서 오른쪽으로 왼쪽 단점 을 매 거 한다. 모든 왼쪽 단점 에 대해 우 리 는 그 를 덮 는 구간 에서 k 대 오른쪽 단점 만 찾 으 면 된다. 따라서 어떤 지원 으로 삭제, 삽입, 조회 k 대 데이터 구 조 를 유지 하면 된다.

#include 
using namespace std;

#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
#define mp make_pair
#define cl(a) memset((a),0,sizeof(a))
typedef pair  pii;
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
// (づ°ω°)づe★------------------------------------------------

#include 
#include 

using namespace __gnu_cxx;
using namespace __gnu_pbds;
  
#define rbset(T) tree,rb_tree_tag,tree_order_statistics_node_update>
typedef pii T;
  
struct RBTree{
    rbset(T) rb;
    void init(){
        rb=rbset(T)();
    }
    void insert(T x){
        rb.insert(x);
    }
    void remove(T x){
        rb.erase(rb.find(x));
    }
    T findkth(int x){ 
        return *rb.find_by_order(x);
    }
    int findRank(T x){
        return rb.order_of_key(x);
    }
}rbtree;

const int maxn = 3e5+5;
pii p[maxn];
setpos;
int n, k;
vectorv;
mapm;

int main(){
	gn(n);gn(k);
	rep(i,1,n){
		gn(p[i].first);gn(p[i].second);
		v.pb(p[i]);
		pos.insert(p[i].first);
	}
	sort(p+1,p+1+n);
	int cnt = 1;
	int ans = 0;
	int ansbg, ansed;
	for(auto it:pos){
		while(cnt <= n && p[cnt].first == it){
			rbtree.insert(mp(-p[cnt].second,++m[p[cnt].second]));
			cnt++;
		}
		auto it2 = rbtree.rb.upper_bound(mp(-it+1,0));
		auto it3 = it2;
		it3++;
		while(it2 != rbtree.rb.end()){
			rbtree.rb.erase(it2);
			it2 = it3;
			if(it3 != rbtree.rb.end()) it3++;
		}
		if(rbtree.rb.size() < k) continue;
		pii kt = rbtree.findkth(k-1);
		int tmpans = -kt.first - it + 1;
		if(tmpans > ans){
			ans = tmpans;
			ansbg = it;
			ansed = -kt.first;
		}
	}
	cout<= ansed){
				tot++;
				cout<