codeforce 754-D(求最大k個重疊區間)


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 hasn discount coupons, the i-th of them can be used with products with ids ranging from li tori, inclusive. Today Fedor wants to take exactlyk coupons with him.

Fedor wants to choose the k coupons in such a way that the number of such productsx that all coupons can be used with this productx 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 andk (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 integersli andri ( - 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 integersp1, p2, ..., pk (1 ≤ pi ≤ n) — the ids of the coupons which Fedor should choose.

If there are multiple answers, print any of them.

Example
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 are31 products in total.

In the second example, no product can be bought with two coupons, that is why the answer is0. Fedor can choose any two coupons in this example.

  題目大意:某人有n張優惠券,每張優惠券都有一個優惠區間,比如說某張優惠券能優惠1~25號商品(1~25號為商品編號,一共25個商品),現在需要從n張優惠劵中選出k張,保證這k張能重疊優惠到最大商品區間來使這個區間中的商品經過多次優惠以至價格最低。本質就是從n個區間中選k個區間。

題解:這題我一開始用到暴力,復雜度為(n*(n-1))/2,沒想到超時了,我的暴力思想是先對所給的區間左值排好序,從第k個商品開始,掃描前面商品,保留最小右值與第k個左值,保留所得區間,然后再從k+1個商品開始,掃描前面的商品,以此下去,不斷改變最大區間,好像在第八組的時候超時了。參考別的博主用的是優先隊列,受益匪淺。

使用優先隊列,也是先對左值排序,選定k個商品,按右值從小到大進隊,求重疊區間,然后出隊(此時出隊肯定是右值最小的,它出隊對於后面區間的沒有影響,因為后面區間左值大於它,重疊的區間不會比它大,自己考慮),不斷更新最大值,這樣只掃描了一遍就得出結果,在輸出時也要注意,自己考慮技巧吧。下面是優先隊列ac代碼。

#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef struct point{
int start;
int end;
int id;
}pt;
pt p[300005];
bool cmp (pt a,pt b)
{
return a.start<b.start;
}
int main ()
{
int n,k;
int i,j,value,flag,l;
while (scanf ("%d%d",&n,&k)!=EOF)
{
priority_queue<int,vector<int>,greater<int> >q;\\隊列申請在while里,放在外面要不斷再把隊列清空。
l=0;
value=0;
for (i=1;i<=n;i++)
{
cin>>p[i].start>>p[i].end;
p[i].id=i;
}
sort (p+1,p+n+1,cmp);
for (i=1;i<=n;i++)
{
l=p[i].start;
q.push(p[i].end);
while (q.size()>k)
q.pop();
if (q.size()==k)
{
if (q.top()-l+1>value)
{
value=q.top()-l+1;
flag=i;
}
}
}
cout<<value<<endl;
if (value==0)
{
for (i=1;i<=k;i++)
{
if (i!=1)
cout<<" ";
cout<<i;
}
cout<<endl;
}
else
{
for (i=1;i<=flag;i++)\\要好好考慮,是個技巧。
if (p[i].end-p[flag].start+1>=value)
cout<<p[i].id<<" ";
cout<<endl;
}
}
return 0;
}





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