Codeforces Round #538 (Div. 2) Yet Another Array Partitioning Task(贪心)
链接:http://codeforces.com/problemset/problem/1114/B
来源:Codeforces
Aay b is called to be a subarray of a if it forms a continuous subsequence of a, that is, if it is equal to al, al+1, …, ar for some l,r.
Suppose m is some known constant. For any array, having m or more elements, let’s define it’s beauty as the sum of m largest elements of that array. For example:
- For array x=[4,3,1,5,2] and m=3, the 3 largest elements of x are 5, 4 and 3, so the beauty of x is 5+4+3=12.
- For array x=[10,10,10] and m=2, the beauty of x is 10+10=20.
You are given an array a1,a2,…,an, the value of the said constant m and an integer k. Your need to split the array a into exactly k subarrays such that:- Each element from a belongs to exactly one subarray.
- Each subarray has at least m elements.
- The sum of all beauties of k subarrays is maximum possible.
Input
The first line contains three integers n, m and k (2≤n≤2⋅105, 1≤m, 2≤k, m⋅k≤n) — the number of elements in a, the constant m in the definition of beauty and the number of subarrays to split to.
The second line contains n integers a1,a2,…,an (−109≤ai≤109).
Output
In the first line, print the maximum possible sum of the beauties of the subarrays in the optimal partition.
In the second line, print k−1 integers p1,p2,…,pk−1 (1≤p1
- All elements with indices from 1 to p1 belong to the first subarray.
- All elements with indices from p1+1 to p2 belong to the second subarray.
- ….
- All elements with indices from pk−1+1 to n belong to the last, k-th subarray.
If there are several optimal partitions, print any of them.
Examples
- Input
9 2 3
5 2 5 2 4 1 1 3 2- Output
21
3 5- Input
6 1 4
4 1 3 2 2 3- Output
12
1 3 5- Input
2 1 2
-1000000000 1000000000- Output
0
1
此题是说将数组a(长度为n)分成k组,每一组选出m个最大的数,将这m*k个数相加能够形成的最大值,输出这个最大值,然后输出每一组的最后一个数的下标(最后一组不输出).这道题我们可以用贪心的方法,对数组排序先选出m*k个最大数的数,对他们的位置做标记.那么我们在分组的时候,每次找到两个被标记的数时我们就找到了一组,直到数组的元素全部找完为止.
#include Node;
typedef long long LL;
const int Max_n=2*100005;
int vis[Max_n],ans[Max_n];
struct Node{
int pos,val;
bool operator <(const Node &n) const{
return val==n.val?pos>n.pos:val>n.val;
}
}a[Max_n];
int main(){
memset(vis,0,sizeof(vis));
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i].val);
a[i].pos=i;
}
LL sum=0;
sort(a+1,a+n+1);
for(int i=1;i<=m*k;i++){//找到m*k个数并且对他们的位置做标记
vis[a[i].pos]=1;
sum+=a[i].val;
}
int t=0,num=1;
for(int i=1;i<=n;i++){
if(vis[i]){//如果此位置被标记过
t++;
if(t==m){//找到了一组,并标记这组的最后一个元素的位置
t=0;
ans[num++]=i;
}
}
}
printf("%lld\n",sum);
for(int i=1;i<k;i++)
printf("%d%c",ans[i],i==k?'\n':' ');
return 0;
}