hdu-1074-Doing Homework-状态压缩dp-java

Problem Description

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test, 1 day for 1 point. And as you know, doing homework always takes a long time. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework).

Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.

 

Output

For each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.

 

Sample Input

    
    
    
    

     
     
     
     2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     2
Computer
Math
English
3
Computer
English
Math


     
     
     
     

      
      
      
      

       
       
       
       Hint
      
      
      
      
In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the 
word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.

     
     
     
     
     
     
     
      
    
    
    
    

 

Author

Ignatius.L

解题思路：
                这是我状态压缩的第一道题 emm 本题是压缩作业的全排列做还是不做 对应的就是01字符串 也就是一个十进制数
               最后求的是 n个1组成的二进制状态 也就是所有作业都完成 也就是求dp[2^n-1] 下面是代码

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class hdu1074 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner scanner = new Scanner(System.in);
        int q = scanner.nextInt();
        for (int p = 0; p < q; p++) {
            int n = scanner.nextInt();
            HomeWork [] subs = new HomeWork[n];
            for (int i = 0; i < n; i++) {
                subs[i] =  new HomeWork(scanner.next(), scanner.nextInt(), scanner.nextInt());
            }
            int max = (int) (Math.pow(2, n) - 1);
            Node [] dp = new Node[max+1];
            dp[0] = new Node(0, 0, ""); //dp[0]表示一个作业都没做
            for (int i = 0; i < dp.length; i++) { //第一层循环 就是枚举所有 的01排列
                for (int j = 0; j < n; j++) { //第二层循环 枚举n门作业 用来计算 我在当前状态做了第j门作业 后跳转到另外一个状态的最优值
                    int indexWork = 1 << j;  //左移j位 模拟第j门作业在二进制串中的位置
                    if ((i & indexWork) == 0) { //判断当前的状态能不能做第j门作业
                        int later = i | indexWork; //如果能作第j门作业 那么跳转到的状态是later
                        if (dp[later] == null) { //如果跳转的状态没值 直接赋值就可以了
                            int t = dp[i].t + subs[j].time;
                            int o = dp[i].o + (t > subs[j].deadline ? t - subs[j].deadline : 0);
                            dp[later] = new Node(t, o, dp[i].w+subs[j].subject+" ");
                        }else {//如果有值 进行比较找到罚时最小的然后更新
                            int t = dp[i].t + subs[j].time;
                            int o = dp[i].o + (t > subs[j].deadline ? t - subs[j].deadline : 0);
                            if (o < dp[later].o) {
                                dp[later] = new Node(t, o, dp[i].w+subs[j].subject+" ");
                            }
                        }
                    }
                }
            }
            System.out.println(dp[max].o);
            String [] sum = dp[max].w.split(" ");
            for (int i = 0; i < sum.length; i++) {
                System.out.println(sum[i]);
            }
        }
    }

}
//dp点
class Node {
    int t; //总用时
    int o; //罚时
    String w; //路径
    public Node(int t, int o, String w) {
        // TODO Auto-generated constructor stub
        this.t = t;
        this.o = o;
        this.w = w;
    }
}
//作业
class HomeWork{
    String subject; //学科
    int deadline;   //截止时间
    int time;       //完成时间
    public HomeWork(String subject, int deadline, int time) {
        // TODO Auto-generated constructor stub
        this.subject = subject;
        this.deadline = deadline;
        this.time = time;
    }
}