Doing Homework

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6299 Accepted Submission(s): 2708

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

/*

读完题目后想的和此博客人想法大致一样
 http://www.cnblogs.com/Kenfly/archive/2011/03/30/2000364.html

可是，感觉有一个问题，因为担心会出现 a,b都可以到c 虽然a到c的花费大
，但是a可能时间短，因为此时间对后面有影响，a可能后面会反败为胜呢？

感觉这样不好处理，百度却找不到解答，最后自己想通了，原来既然状态相同，那么花费的时间是一样的，
这样疑惑顿时没有了




*/


#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
         
           #include
          
            #include
            #define L(x) (x<<1) #define R(x) (x<<1|1) #define MID(x,y) ((x+y)>>1) #define bug printf(hihi ) #define eps 1e-8 typedef __int64 ll; using namespace std; #define INF 0x3f3f3f3f #define N 16 struct stud{ int pre,time;//当前状态是由哪一个点来的，当前状态时间， int score;//当前扣得分数 }dp[1<<16]; struct ha{ string name; int d,len; }home[N]; int n; void show(int st) { if(st==0) return ; show(st^(1<
            
             home[j].d) score=time-home[j].d; score+=dp[i].score; if(dp[to].score>score) { dp[to].score=score; dp[to].pre=j; dp[to].time=time; // printf(%d: %d %d %d ,to,to^(1<
             
              >home[i].name>>home[i].d>>home[i].len;//字典序输入，无需排序 } DP(); } return 0; }


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