【Codeforces 1114B】Yet Another Array Partitioning Task

【链接】 我是链接,点我呀:)

【题意】

让你把数组分成k个连续的部分
使得每个部分最大的m个数字的和最大

【题解】

把原数组降序排序
然后选取前m*k个数字打标记
然后对于原数组
一直贪心地取
直到这个区间选了m个打标记的数字为止。
然后就划分一个区间>_

【代码】

import java.io.*;
import java.util.*;
public class Main {
	static int N = (int)2e5;
	static InputReader in;
	static PrintWriter out;
	static class Pair implements Comparable<Pair>{
		int x,id;
		public Pair(int x,int id) {
			this.x = x;this.id = id;
		}
		@Override
		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			return o.x-this.x;
		}
	}
	public static void main(String[] args) throws IOException{
		//InputStream ins = new FileInputStream("E:\\rush.txt");
		InputStream ins = System.in;
		in = new InputReader(ins);
		out = new PrintWriter(System.out);
		//code start from here
		new Task().solve(in, out);
		out.close();
	}
	static class Task{
		public void solve(InputReader in,PrintWriter out) {
			int n,m,k;
			int []a = new int [N+10];
			int []tag = new int [N+10];
			Pair []b = new Pair[N+10];
			n = in.nextInt();m = in.nextInt();k = in.nextInt();
			for (int i = 1;i <= n;i++) a[i] = in.nextInt();
			for (int i = 1;i <= n;i++) {
				b[i] = new Pair(a[i],i);
			}
			Arrays.sort(b, 1,n+1);
			long ans1 = 0;
			for (int i = 1;i <=m*k;i++) {
				tag[b[i].id]= 1;
				ans1 = ans1 + b[i].x;
			}
			out.println(ans1);
			int cnt = 0;
			int cnt2 = 0;
			for (int i = 1;i <= n;i++){
				if (tag[i]==1) {
					cnt++;
					if (cnt==m) {
						cnt = 0;
						out.print(i+" ");
						cnt2++;
						if (cnt2==k-1){
							return;
						}
					}
				}
			}
		}
	}
	static class InputReader{
		public BufferedReader br;
		public StringTokenizer tokenizer;
		public InputReader(InputStream ins) {
			br = new BufferedReader(new InputStreamReader(ins));
			tokenizer = null;
		}
		public String next(){
			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
				try {
				tokenizer = new StringTokenizer(br.readLine());
				}catch(IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}