题解 | #餐馆#贪心 最后一个用例在超时的边缘徘徊

import java.util.*;

public class Main {
    static boolean[] used;
    static long res;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(); // 桌子数量
        int m = sc.nextInt(); // 顾客数量
        used = new boolean[n];
        int[] tables = new int[n];
        for (int i = 0; i < n; i++) {
            tables[i] = sc.nextInt();
        }
        int[][] nums = new int[m][2]; // 顾客人数和消费
        for (int i = 0; i < m; i++) {
            nums[i][0] = sc.nextInt();
            nums[i][1] = sc.nextInt();
        }
        //消费高的人少的先来
        Arrays.sort(nums, (x, y)-> {
            return x[1] != y[1] ? y[1] - x[1] : x[0] - y[0];
        });
        //桌子小的先使用
        Arrays.sort(tables);
        for (int j = 0; j < m; j++) {
            int people = nums[j][0];
            int money = nums[j][1];
            dfs(tables,n,people,money);
        }
        System.out.println(res);
    }

    private static void dfs(int[] tables,int n,int people,int money) {
        for (int i=0; i<n; i++) {
            if (!used[i]) {
                if (tables[i] >= people) {
                    res +=(long)money;
                    used[i] = true;
                    break;
                } 
            }
        }
    }
}