DP-求最大值求最优解

版权声明：fromZjy QQ1045152332 https://blog.csdn.net/qq_36762677/article/details/84349798

Description

分配了8个任务,时间不冲突的情况下,挣得工资最多

Solution

1.求OPT(i)
如果选了第i个,公式为第i个能挣得钱数加上上一个能选的OPT
Vi+OPT(prev(i))
如果没选第i个,公式就是求上一个能赚的最多的钱数
OPT(i-1)
解释
OPT(i)做第i个任务,能赚多少钱
prev(i)做第i个任务,上一个能做的任务的位置

2.求出前置任务

3.

import java.util.Scanner;

class job {//任务类
    int start;
    int end;
    int payment;
}

public class DPMaxSalary {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int count = sc.nextInt();
        //保存输入信息到jb对象数组
        job[] jb = new job[count];
        for (int i = 0; i < count; i++) {
            jb[i] = new job();
        }
        for (int i = 0; i < count; i++) {
            jb[i].start = sc.nextInt();
            jb[i].end = sc.nextInt();
            jb[i].payment = sc.nextInt();
        }
        //==========录入前置节点==================
        int pre[] = new int[count];
        pre[0] = 0;
        for (int i = 1; i < count; i++) {
            for (int j = i - 1; j >= 0; j--) {
           	 //一直没进if的自动初始化为0
                if (jb[i].start >= jb[j].end) {
                    pre[i] = j + 1;
                    break;
                }
            }
        }
		//==========最优解==================
		int opt[] = new int[count];
        opt[0] = jb[0].payment;

        for (int i = 1; i < count; i++) {
            if (pre[i] == 0) {
                opt[i] = jb[i].payment > opt[i - 1] ? jb[i].payment : opt[i - 1];
            } else {
                opt[i] = opt[i - 1] > opt[pre[i - 1]] + jb[i].payment ? opt[i - 1] : opt[pre[i - 1]] + jb[i].payment;
            }
        }
        for (int i = 0; i < count; i++) {
            System.out.println("第" + (i + 1) + "个点的最优解是:" + opt[i]);
        }
    }
}