# 题目

## 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.

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

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.

# 题解

`dp[i]` 表示当前处于 `i` 状态

`dp[NewState].min > dp[i].min + ReduceScore`

(由于本身科目是按照字典序的,并且只有严格小于会更新状态,因此最后答案必然是字典序最小的)

# 代码

```/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;

const int INF = 0x7FFFFFFF / 2;
const int maxn = 16;
const int maxstate = (1 << 15);

struct Node {
string Name;
int Time;
};
struct Node2 {
int min;
int time;
int last;
};
Node Homework[maxn];
Node2 dp[maxstate];
stack<string> s;

int SetI(int num,int i,bool flag) {
if(flag)
return num | (1 << i);
else
return num & (0 << i);
}
bool GetI(int num,int i) {
return num >> i & 1;
}
int GetChange(int num1,int num2) {
int n = num1 ^ num2;
int i = 0;
while(n) {
if(n & 1)
return i;
n >>= 1;
i++;
}
return 0;
}

void Do() {
int n;
cin >> n;

int sum = 0;
int mt = 0;

for(int i = 0;i < n;i++)
cin >> Homework[i].Name >> Homework[i].Deadline >> Homework[i].Time;

dp.min = 0;
for(int i = 1;i <= (1 << n) - 1;i++)
dp[i].min = INF;

for(int i = 0;i < (1 << n) - 1;i++) {
for(int j = 0;j < n;j++)
if(!GetI(i,j)) {
int NewState = SetI(i,j,1);
int Time = dp[i].time + Homework[j].Time;
int ReduceScore = (Time > Homework[j].Deadline) ? (Time - Homework[j].Deadline) : 0;
if(dp[NewState].min > dp[i].min + ReduceScore) {
dp[NewState].min = dp[i].min + ReduceScore;
dp[NewState].time = Time;
dp[NewState].last = i;
}
}
}

cout<<dp[(1 << n) - 1].min<<endl;

int k = (1 << n) - 1;
while(k) {
int pos = GetChange(k,dp[k].last);
s.push(Homework[pos].Name);
k = dp[k].last;
}

while(!s.empty()) {
cout << s.top() << endl;
s.pop();
}
return;
}
int main() {
cin.tie(0);
std::cin.sync_with_stdio(false);

int T;
cin >> T;
while(T--)
Do();

return 0;
}
```