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HDU 1074 Doing Homework（狀壓dp）

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HDU 1074 Doing Homework（狀壓dp）

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

/*

讀完題目後想的和此博客人想法大致一樣
 http://www.cnblogs.com/Kenfly/archive/2011/03/30/2000364.html

可是，感覺有一個問題，因為擔心會出現 a,b都可以到c 雖然a到c的花費大
，但是a可能時間短，因為此時間對後面有影響，a可能後面會反敗為勝呢？

感覺這樣不好處理，百度卻找不到解答，最後自己想通了，原來既然狀態相同，那麼花費的時間是一樣的，
這樣疑惑頓時沒有了




*/


#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;//當前狀態是由哪一個點來的，當前狀態時間，
  int score;//當前扣得分數
}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;//字典序輸入，無需排序
        }
        DP();
    }
    return 0;
}