[https://atcoder.jp/contests/abc234/tasks/abc234_d Prefix K-th Max] 最小堆实现

Problem Statement

Given are a permutation P=(P_1,P_2,\ldots,P_N)P=(P1​,P2​,…,PN​) of (1,2,\ldots,N)(1,2,…,N) and a positive integer KK.

For each i=K,K+1,\ldots,Ni=K,K+1,…,N, find the following.

• The KK-th greatest value among the first ii terms of PP.

Constraints

• 1 \leq K \leq N \leq 5 \times 10^51≤K≤N≤5×105
• (P_1,P_2,\ldots,P_N)(P1​,P2​,…,PN​) is a permutation of (1,2,\ldots,N)(1,2,…,N).
• All values in input are integers

解法：

这就是1个TOPK问题。

java的最小堆、最大堆可以使用PriorityQueue<Integer>

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.PriorityQueue;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int N,K;
        String[] words = reader.readLine().split("\\s+");
        N = Integer.parseInt(words[0]);
        K = Integer.parseInt(words[1]);

        int[] P = new int[N];
        words = reader.readLine().split("\\s+");
        for(int i=0;i<N;i++){
            P[i] = Integer.valueOf(words[i]);
        }
        PriorityQueue<Integer> pq = new PriorityQueue<>(K);
        for(int i =0;i<K;i++){
            pq.add(P[i]);
        }
        StringBuilder sb = new StringBuilder();
        sb.append(pq.peek()+"\n");
        for(int i=K;i<N;i++){
            int val = P[i];
            if( val > pq.peek()){
                pq.poll();
                pq.add(val);
            }
            sb.append(pq.peek()+"\n");
//            System.out.println(pq.peek());
        }
        System.out.println(sb);
        reader.close();
    }

}