题面

核心思想

定义一个三维数组 dp[i][t][h] 表示前i个事件在时间剩余t精力剩余h的最大快乐值
每个事件考虑分享or不分享，然后取最大值

代码

import java.util.*;

public class Main {

    public static void main(String[] args) {
        final long MOD = (long) (1e9 + 7);
        Scanner scanner = new Scanner(System.in);
        int n  = scanner.nextInt();
        int T = scanner.nextInt(), H = scanner.nextInt();
        // 前i个事件在时间剩余t精力剩余h的最大快乐值
        long[][][] dp = new long[n + 1][T + 1][H + 1];
        int[][] event = new int[n][3];
        for(int i = 0; i < n; i++){
            int x = scanner.nextInt();
            int y = scanner.nextInt();
            int z = scanner.nextInt();
            event[i][0] = x;
            event[i][1] = y;
            event[i][2] = z;
        }

        for(int i = 1; i <=n; i++){
            for(int t = 0; t <= T; t++){
                for(int h = 0; h <= H; h++){
                    //剩余时间
                    int leftT = t - event[i - 1][0];
                    //剩余精力
                    int leftH = h - event[i - 1][1];

                    //不分享事件i
                    dp[i][t][h] = dp[i - 1][t][h];

                    //能够分享事件i 则取最大值
                    if(leftT >= 0 && leftH >= 0)
                        dp[i][t][h] = Math.max(dp[i][t][h], dp[i - 1][leftT][leftH] + event[i - 1][2]);
                }
            }
        }
        System.out.println(dp[n][T][H]);
    }
}