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

題意是要求你完成一寫作業，每份作業都有自己的截止日期和所需時間，如果晚交一天就要扣一分，問做完這些作業最少需要扣幾分，這道題感覺得NP難題，好在作業數量只有16個，可以用狀壓DP進行狀態壓縮，複雜度為O（2^n*n^2），狀壓DP的優化就是對記憶體空間的優化，下面是自己的程式碼：

#include<bits/stdc++.h>
using namespace std;
using LL=int64_t;
const int INF=0x3f3f3f3f;
const int maxn=1<<16;
int dp[maxn],dead[maxn],pre[maxn];
int n;
struct Node {
    string s;
    int cost,dead;
}ans[20];

void print(int num) {
    if(num==0) return ;
    print(pre[num]);
    num-=pre[num];//此時的num為一個i位是1其他位都是0的數，減去以後表示此時做這個作業是最優解
    for(int i=0;i<n;i++) {
        if(num&1<<i) cout<<ans[i].s<<endl;
    }
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--) {
        cin>>n;
        for(int i=0;i<n;i++) cin>>ans[i].s>>ans[i].dead>>ans[i].cost;
        int sum=1<<n;
        for(int i=1;i<sum;i++) dp[i]=INF;
        dp[0]=0;
        memset(dead,0,sizeof(dead));
        memset(pre,0,sizeof(pre));
        for(int i=1;i<sum;i++) {
            for(int j=0;j<n;j++) {
                if((i&1<<j)==0) continue;//如果i對應的j位是0表示沒做這個作業
                int k=i-(1<<j);//表示沒做作業這個的情況
                dead[i]=dead[k]+ans[j].cost;
                int cost=dead[i]>ans[j].dead?dead[i]-ans[j].dead:0;
                if(dp[k]+cost<=dp[i]) {
                    dp[i]=dp[k]+cost;
                    pre[i]=k;
                }
            }
        }
        cout<<dp[sum-1]<<endl;
        print(sum-1);
    }
    return 0;
}