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
题意：
有n门课程作业，每门作业的截止时间为D,需要花费的时间为C，若作业不能按时完成，每超期1天扣1分。这n门作业按课程的字典
序先后输入。
问完成这n门作业至少要扣多少分，并输出扣分最少的做作业顺序。
PS:达到扣分最少的方案有多种，请输出字典序最小的那一组方案。
分析：
因为最多只有15门课程，可以使用二进制来表示所有完成的状况。
例如5，二进制位101，代表第一门和第三门完成了，第二门没有完成，那么我们可以枚举1~1<<n便可以得出所有的状态。
然后对于每一门而言，其状态是cur = 1<<i，我们看这门在现在的状态j下是不是完成，可以通过判断cur&j是否为1来得到。
当得出cur属于j状态的时候，我们便可以进行dp了，在dp的时候要记录路径，方便之后的输出。
代码：
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
const int MAXN=1<<16;
int n;
struct Node
{
char name[109];//课程名
int endTime;//结束的时间
int cost;//完成需要的时间
}course[20];
int vis[MAXN];
struct Dp
{
int cost;//时间
int pre;//前一状态
int reduced;//损失的分数
} dp[MAXN];
void out(int status)
{
int curJob=dp[status].pre^status;//当前那项工作被做的状态
int curid=0;
curJob>>=1;
while(curJob)//获得当前做的这个工作的编号
{
curid++;
curJob>>=1;
}
if(dp[status].pre!=0)//没到第一项工作就一直往前递归
{
out(dp[status].pre);
}
printf("%s\n",course[curid].name);//把这项工作的名称输出来
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(vis,0,sizeof(vis));
scanf("%d",&n);
for(int i=0; i<n; i++)
scanf("%s%d%d",&course[i].name,&course[i].endTime,&course[i].cost);
dp[0].cost=0;
dp[0].pre=-1;
dp[0].reduced=0;
vis[0]=1;
for(int j=0; j<1<<n; j++) //遍历所有的状态
{
for(int work=0; work<n; work++)
{
int cur=1<<work;
if((j&cur)==0)//第work项作业没有做
{
int curTemp=cur|j;//在j的状态下又做了第work个工作
int day=dp[j].cost+course[work].cost;//需要花费的时间
dp[curTemp].cost=day;
int reduce=day-course[work].endTime;
if(reduce<0) reduce=0;//有足够的时间来完成这项作业
reduce+=dp[j].reduced;//得加上j状态下浪费得时间
if(vis[curTemp])//已经访问过
{
if(dp[curTemp].reduced>reduce)//并且扣除的分数要更新
{
dp[curTemp].reduced=min(reduce,dp[curTemp].reduced);
dp[curTemp].pre=j;
}
}
else//没有访问过
{
vis[curTemp]=1;
dp[curTemp].reduced= reduce;
dp[curTemp].pre=j;
}
}
}
}
printf("%d\n",dp[(1<<n)-1].reduced);
out((1<<n)-1);//递归输出
}
return 0;
}
