D. Fedor and coupons(思路)

D. Fedor and couponstime limit per test4 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAll 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!InputThe 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.OutputIn 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.ExamplesinputCopy

4 2
1 100
40 70
120 130
125 180

output

31
1 2

inputCopy

3 2
1 12
15 20
25 30

output

0
1 2

inputCopy

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

output

21
3 4

NoteIn 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.

题意：给出n个区间，求m个区间的最大覆盖，并输出覆盖的区间编号
现将n个区间的起始时间从小到大排序，然后优先队列（结束时间从小到大）维护m个区间，这样保证每次选的都是最小的r，然后用这个最小的r减去这m个区间的最大l，此时求出的即为覆盖m次的区间,刷新最大值即可#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
const int N=3e5+50;
struct node
{
int l,r,num;
}e
;
priority_queue<int,vector<int>,greater<int> >q;
bool cmp(node a,node b)
{
return a.l<b.l;
}
int main()
{
int n,k,maxx=0,xx,yy,m=0;
cin>>n>>k;
for(int i=1;i<=n;i++)
{
cin>>e[i].l>>e[i].r;
e[i].num=i;
}
sort(e+1,e+1+n,cmp);
for(int i=1;i<=n;i++)
{
int x=e[i].l,y=e[i].r;
q.push(y);
if(q.size()>k)q.pop();
int len=q.top()-x+1;
if(q.size()==k&&maxx<len)
{
maxx=len;
xx=x;
yy=q.to
cccd
p();
}
}
printf("%d\n",maxx);
if(maxx==0)xx=INF,yy=-INF;
for(int i=1;i<=n&&m<k;i++)
if(e[i].l<=xx&&e[i].r>=yy)
m++,printf("%d ",e[i].num);

return 0;
}