Fedor and coupons CodeForces - 754D (优先队列，贪心）

http://codeforces.com/problemset/problem/754/D

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 l_i to r_i, 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!

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 l_i and r_i ( - 10⁹ ≤ l_i ≤ r_i ≤ 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, ..., p_k (1 ≤ p_i ≤ n) — the ids of the coupons which Fedor should choose.

If there are multiple answers, print any of them.

Example

Input

Output

31
1 2

Input

Output

0
1 2

Input

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.

题意：每个商品对应一个编号，有n张优惠券，每张优惠券可以可以优惠 left到 right区间中的商品。现在选择k张优惠券，问能优惠k次的商品最多有多少个，怎么选择优惠券？

思路：区间先排序，此题应该先将区间以L的大小排序，L小的排在前面，R排不排不影响。

排完序后，你所选的k个区间的重叠部分由他们中最小的R，和最大的L确定，L已经是排好序的，所有的L都大于等于前面的L，所以L不用再考虑，当前的L就是最大的。最小的R用优先队列来维护，在你选择的k个区间中，拿出最小的R，比较是否比原来的解更优（R-L+1），更优则跟新。每次将k个区间的最小的R删除（删除最小的R为最优，留下的R越大越好），所选的重叠的区间可以有两个元素确定（区间长度和L），所以每次更新答案，将最优的L值也保存下来。

代码：

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#define N 300500
#define mem(a,b) memset(a,b,sizeof(a));
using namespace std;
struct node
{
    int l,r,id;
} a[N];
bool cmp(node b,node c)
{
    if(b.l==c.l) return b.r<c.r;
    return b.l<c.l;
}
priority_queue<int, vector<int>, greater<int> > q;//优先队列，小的优先，不会优先队列的可以自己百度一下
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++)
    {
        scanf("%d%d",&a[i].l,&a[i].r);
        a[i].id=i;//存一下区间的编号，因为一会要排序。
    }
    sort(a+1,a+1+n,cmp);
    int ans=0,t=0;//ans表示最大的区间长度，t表示最优解的区间左端点
    for(int i=1; i<=k-1; i++)
        q.push(a[i].r);//先选k-1个区间，将R放进队列
    for(int i=k; i<=n; i++)//每次选第i个，删掉R最小的那个，维护队列中有k-1个区间
    {
        q.push(a[i].r);
        int w=q.top();//q.top()是返回队列中r最小值，
        if(ans<w-a[i].l+1)
        {
            ans=w-a[i].l+1;//更新答案
            t=a[i].l;//更新答案的左端点
        }
        q.pop();//删除最小的r
    }
    if(ans==0)//答案为零，这样输出
    {
        printf("0\n");
        for(int i=0; i<k; i++)
        {
            if(i) printf(" ");
            printf("%d",i+1);
        }
        printf("\n");
        return 0;
    }
    printf("%d\n",ans);
    int flag=0;
    for(int i=1; i<=n; i++)
    {
        if(k&&a[i].l<=t&&a[i].r>=ans+t-1)//所有包含[t,t+ans-1]这段区间的区间都可以选。
        {
            k--;//共输出k个
            if(flag) printf(" ");
            printf("%d",a[i].id);
            flag=1;
        }
    }
    printf("\n");
}

Fedor and coupons CodeForces - 754D (优先队列，贪心）

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