Fedor and coupons

time limit per test 4 second memory limit per test 256 megabytes
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 n discount 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 k coupons 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!

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Input

The first line contains two integers n and k (1≤k≤n≤3·10^5) — 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 (-10^9≤li≤ri≤10^9) — 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.

Sample Input

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2
3
4
5
4 2
1 100
40 70
120 130
125 180

Sample Output

1
2
31
1 2

题解

心疼自己宛如一个智障
这真的是一个好题
思路有点像LIS中nlogn算法中那种动态扩张的思想
其实应该算贪心吧
但确实
很厉害
看了别人的题解才知道自己还是太菜了
先把所有区间按左起点从小到大排,这样我们后面每次更新时都能严格保证每次公共区间的起点是最后加进去的那个区间的左起点。
然后用一个优先队列保存k个区间,按右起点从小到大排,这样我们能保证每次弹出的是右起点最小的值。
最后注意先排序时会打乱下标,要输出真正的下标
还是不会就去http://codeforces.com/blog/entry/49613 看Saat的回答吧

AC code:(不包含输入类)

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import java.io.*;  
import java.util.*;
public class Main {
static class Coupons implements Comparable<Coupons>{
int left;
int right;
int index;
Coupons(int a,int b,int c){
this.left=a;
this.right=b;
this.index=c;
}
@Override
public int compareTo(Coupons o) {
if(this.right<o.right) return -1;
else return 1;
}

}
static class LeftComparable implements Comparator<Coupons>{
@Override
public int compare(Coupons a, Coupons b) {
if(a.left<b.left)
return -1;
else return 1;
}
}
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
int k=sc.nextInt();
Coupons []cou=new Coupons[n];
for(int i=0;i<n;i++){
cou[i]=new Coupons(sc.nextInt(),sc.nextInt(),i);
}
Arrays.sort(cou,new LeftComparable());
PriorityQueue<Coupons>queue=new PriorityQueue<Coupons>();
int result=Integer.MIN_VALUE;
int left=0;
int right=0;
//ArrayList<Integer>list=new ArrayList<Integer>(); 每次记录会超时
for(int i=0;i<n;i++){
queue.add(cou[i]);
if(queue.size()>k) queue.poll();
if(queue.size()==k){
if(result<queue.peek().right-cou[i].left+1){
result=queue.peek().right-cou[i].left+1;
left=cou[i].left;
right=queue.peek().right;
}
}
}
//System.out.println(result);
if(result<=0){
System.out.println(0);
for(int i=1;i<=k;i++){
if(i==n) System.out.print(i);
else System.out.print(i+" ");
}
}
else{
System.out.println(result);
int count=k;
ArrayList<Integer>al=new ArrayList<Integer>();
for(int i=0;i<n;i++){
if(count>0){
if(cou[i].left<=left&&cou[i].right>=right){
al.add(cou[i].index+1);
count--;
}
}
else break;
}
Collections.sort(al);
for(int i=0;i<al.size();i++){
if(i==al.size()-1)
System.out.print(al.get(i));
else
System.out.print(al.get(i)+" ");
}
}
}

}

}