Doing Homework

位操作实现技巧： 
如果要获得第i位的数据，判断((data&(0x1<<i))==0),若真，为0，假，为1； 
如果要设置第i位为1，data=(data|(0x1<<i)); 
如果要设置第i位为0，data=(data&(~(0x1<<i))); 
如果要将第i位取反，data=(data^(0x1<<i); 

如果要取出一个数的最后一个1(lowbit)：(data&(-data))         （这里利用的是负数取反加1实际上改变的是二进制最低位的1这个性质）

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.

        
 

代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
struct INF{
    char name[106];
    int limit,use;
}a[20];
const int NMAX=1<<15|10;
const int inf=(1<<31)-1;
int time[NMAX],dp[NMAX],path[NMAX];//time 该状态所用的时间 dp 该状态被扣的分
void out(int x)
{
    if(x==0)
        return;
    out(x-(1<<path[x]));
    printf("%s\n",a[path[x]].name);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        memset(time,0,sizeof(time));
        for(int i=0;i<n;i++)
            scanf("%s%d%d",&a[i].name,&a[i].limit,&a[i].use);
        int choice=1<<n;
        for(int i=1;i<choice;i++)
        {
            dp[i]=inf;
            for(int j=n-1;j>=0;j--)
            {
                int num=1<<j;//要增加的第j门
                if(!(i&num))
                    continue;//若i中没有第j门 退出
                int score=time[i-num]+a[j].use-a[j].limit;
                if(score<0)//得分小于0, 则为0
                    score=0;
                if(dp[i]>dp[i-num]+score)//更新
                {
                    dp[i]=dp[i-num]+score;
                    time[i]=time[i-num]+a[j].use;
                    path[i]=j;
                }
            }

        }
        printf("%d\n",dp[choice-1]);
        out(choice-1);
    }
    return 0;
}