Codeforces Round #398(Div. 2)D. Cartons of milk【二分+暴力处理】

D. Cartons of milk

time limit per test

2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Olya likes milk very much. She drinks k cartons of milk each day if she has at least
k and drinks all of them if she doesn't. But there's an issue — expiration dates. Each carton has a date after which you can't drink it (you still can drink it exactly at the date written on the carton). Due to this, if
Olya's fridge contains a carton past its expiry date, she throws it away.

Olya hates throwing out cartons, so when she drinks a carton, she chooses the one which expires the fastest. It's easy to understand that this strategy minimizes the amount of cartons thrown out and lets her avoid it if it's even possible.


Milk. Best before: 20.02.2017.

The main issue Olya has is the one of buying new cartons. Currently, there are
n cartons of milk in Olya's fridge, for each one an expiration date is known (how soon does it expire, measured in days). In the shop that Olya visited there are
m cartons, and the expiration date is known for each of those cartons as well.

Find the maximum number of cartons Olya can buy so that she wouldn't have to throw away any cartons. Assume that Olya drank no cartons today.

Input
In the first line there are three integers n,
m, k (1 ≤ n, m ≤ 106,
1 ≤ k ≤ n + m) — the amount of cartons in Olya's fridge, the amount of cartons in the shop and the number of cartons Olya drinks each day.

In the second line there are n integers
f1, f2, ..., fn (0 ≤ fi ≤ 107) —
expiration dates of the cartons in Olya's fridge. The expiration date is expressed by the number of days the drinking of this carton can be delayed. For example, a
0 expiration date means it must be drunk today,
1 — no later than tomorrow, etc.

In the third line there are m integers
s1, s2, ..., sm (0 ≤ si ≤ 107) —
expiration dates of the cartons in the shop in a similar format.

Output
If there's no way for Olya to drink the cartons she already has in her fridge, print
-1.

Otherwise, in the first line print the maximum number x of cartons which Olya can buy so that she wouldn't have to throw a carton away. The next line should contain exactly
x integers — the numbers of the cartons that should be bought (cartons are numbered in an order in which they are written in the input, starting with
1). Numbers should not repeat, but can be in arbitrary order. If there are multiple correct answers, print any of them.

Examples

Input

3 6 2
1 0 1
2 0 2 0 0 2

Output

3
1 2 3

Input

3 1 2
0 0 0
1

Output

-1

Input

2 1 2
0 1
0

Output

1
1

Note
In the first example k = 2 and Olya has three cartons with expiry dates
0, 1 and
1 (they expire today, tomorrow and tomorrow), and the shop has
3 cartons with expiry date 0 and
3 cartons with expiry date 2. Olya can buy three cartons, for example, one with the expiry date
0 and two with expiry date 2.

In the second example all three cartons Olya owns expire today and it means she would have to throw packets away regardless of whether she buys an extra one or not.

In the third example Olya would drink k = 2 cartons today (one she alreay has in her fridge and one from the shop) and the remaining one tomorrow.

题目大意：

主人公很喜欢喝牛奶，每天会喝K盒，如果不足K盒，就会将所有拥有的都喝了。

不傻的人都知道，牛奶过期了之后肯定就不能喝了，那就只能扔掉。

现在冰箱中有N盒牛奶，超市中有M盒牛奶，如果冰箱中的牛奶喝不完就有浪费的情况出现，输出-1，否则输出能够在超时中购买的牛奶数量的最大值。

并且输出可以购买的牛奶的编号。

思路：

1、首先我们将冰箱中的牛奶按照到期时间按照从小到大排序，很显然我们希望先喝到期时间早的，才能更多的避免浪费。

接下来O（n）暴力判断一下这些牛奶是否都能喝完，如果能，继续进行，否则输出-1.

2、接下来我们按照到期时间从大到小排序，很明显，我们希望购买的牛奶肯定是越晚到期越好。

接下来考虑到这样一点：购买的牛奶越多，越有可能要浪费掉，所以我们知道这里有一个单调性，我们考虑二分购买的牛奶数量。

对于二分出来的中间值mid，我们暴力处理一次判断能否将这N+mid盒牛奶都喝完，如果可以，增加购买数量，否则减少购买数量。

过程维护最后一次可行方案数，输出即可。

Ac代码（搓比代码底层优化没做好，卡边界1900+ms过的）：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
int date,pos;
}b[1000700];
int a[1000700];
int c[2007000];
int output[2000700];

int n,m,k;
int cmp(node a,node b)
{
return a.date>b.date;
}
int judge(int a[],int n)
{
int day=0;
int cnt=0;
for(int i=0;i<n;i++)
{
if(day>a[i])return 0;
cnt++;
if(cnt==k)
{
cnt=0;
day++;
}
}
return 1;
}
int Slove(int mid)
{
int tot=0;
for(int i=0;i<n;i++)c[tot++]=a[i];
for(int i=0;i<=mid;i++)c[tot++]=b[i].date;
sort(c,c+tot);
if(judge(c,tot)==1)
{
for(int i=0;i<=mid;i++)
{
output[i]=b[i].pos;
}
return 1;
}
else return 0;
}
int main()
{
while(~scanf("%d%d%d",&n,&m,&k))
{
for(int i=0;i<n;i++)scanf("%d",&a[i]);
for(int i=0;i<m;i++)scanf("%d",&b[i].date),b[i].pos=i+1;
sort(a,a+n);sort(b,b+m,cmp);
if(judge(a,n)==0)
{
printf("-1\n");
continue;
}
int ans=-1;
int l=0;
int r=m-1;
while(r-l>=0)
{
int mid=(l+r)/2;
if(Slove(mid)==1)
{
l=mid+1;
ans=mid;
}
else
{
r=mid-1;
}
}
printf("%d\n",ans+1);
for(int i=0;i<ans+1;i++)
{
printf("%d ",output[i]);
}
printf("\n");
}
}