codeforces 754D. Fedor and coupons

D. Fedor and coupons
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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!

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 li and ri ( - 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 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.

Examples
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 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。
　然而考完之后发现了，二分个毛线啊！
 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 }

 

posted @ 2017-01-07 01:02  №〓→龙光←  阅读(438)  评论(0)  编辑  收藏  举报