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 xis 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!
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.
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.
4 2
1 100
40 70
120 130
125 180
31
1 2
3 2
1 12
15 20
25 30
0
1 2
5 2
1 10
5 15
14 50
30 70
99 100
21
3 4
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：给出n个区间，要求选出k个区间，是k个区间的并尽可能大
话说无数神犇在我比赛的时候指点了我，但是MDZZ。。。我都没听懂啊，最后yy一个丑陋复杂度的做法。
考虑对于l为第一关键字排序，然后二分答案，可知l是单调递增的，用一个堆维护第k大的r是什么，然后判断区间大小就可以了。
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<vector> 5 #include<cstdlib> 6 #include<cmath> 7 #include<cstring> 8 #include<queue> 9 using namespace std; 10 #define maxn 500010 11 #define llg long long 12 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout); 13 llg n,m,k,mid,anss[maxn],flag; 14 15 struct data 16 { 17 llg quan,wz; 18 bool operator < (const data &a )const 19 { 20 return a.quan<quan; 21 } 22 }; 23 24 priority_queue<data>q; 25 26 struct node 27 { 28 llg l,r,wz; 29 }a[maxn]; 30 31 bool cmp(const node&a,const node&b) {return a.l<b.l;} 32 33 bool check(llg x) 34 { 35 if (!flag) 36 { 37 while (!q.empty()) q.pop(); 38 } 39 else 40 { 41 flag=0; 42 llg tail=0; 43 while (!q.empty()) {anss[++tail]=q.top().wz; q.pop();} 44 } 45 for (llg i=1;i<=n;i++) 46 { 47 data w; 48 w.quan=a[i].r; w.wz=a[i].wz; 49 q.push(w); 50 if (q.size()>k) q.pop(); 51 if (q.size()==k) 52 { 53 if (q.top().quan-a[i].l+1>=x) return 1; 54 } 55 } 56 return 0; 57 } 58 59 int main() 60 { 61 yyj("D"); 62 cin>>n>>k; 63 for (llg i=1;i<=n;i++){ scanf("%lld%lld",&a[i].l,&a[i].r); a[i].wz=i;} 64 sort(a+1,a+n+1,cmp); 65 llg l=0,r=(llg)20000000000; 66 llg ans=0; 67 while (l<=r) 68 { 69 mid=(l+r)/2; 70 if(check(mid)) {ans=mid; l=mid+1; flag=1;} else {r=mid-1;} 71 } 72 cout<<ans<<endl; 73 sort(anss+1,anss+k+1); 74 if (ans!=0) 75 { 76 for (llg i=1;i<=k;i++) printf("%lld ",anss[i]); 77 } 78 else 79 { 80 for (llg i=1;i<=k;i++) printf("%lld ",i); 81 } 82 return 0; 83 }