算法分析

单调队列

以i结尾的f[i]，滑动窗口的区间是[i - m,i - 1]，单调队列维护的是该区间的最小值，由于滑动窗口不包含i，因此f[i]需要在while上方进行更新

时间复杂度

Java 代码

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    static int N = 2 * 100000 + 10;
    static int[] q = new int[N];
    static int[] a = new int[N];
    static int n,m;
    static int[] f = new int[N];
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] s1 = br.readLine().split(" ");
        n = Integer.parseInt(s1[0]);
        m = Integer.parseInt(s1[1]);
        String[] s2 = br.readLine().split(" ");
        for(int i = 1;i <= n;i ++) 
        {
            a[i] = Integer.parseInt(s2[i - 1]);
        }

        int hh = 0,tt = -1;
        int res = 0x3f3f3f3f;
        for(int i = 0; i <= n;i ++)
        {
            if(hh <= tt && q[hh] < i - m) hh ++;
            f[i] = f[q[hh]] + a[i];
            while(hh <= tt && f[q[tt]] >= f[i]) tt --;
            q[++ tt] = i;

            if(i >= n - m + 1) res = Math.min(res,f[i]);
        } 
        System.out.println(res);
    }
}