HDU 1074 Doing Homework（状压DP+记录路径）

原题地址

Doing Homework

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8336 Accepted Submission(s): 3849

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

23Computer 3 3English 20 1Math 3 23Computer 3 3English 6 3Math 6 3

Sample Output

2ComputerMathEnglish3ComputerEnglishMath
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.
 

题意是说，一共给你按名字的字典序排列的n门功课。每一门功课有名字，最晚完成时间，完成所需要的时间。

一门功课的完成时间每超过最晚完成时间1天扣一分，请问把所有功课完成的最少扣分数的多少，并且输出这种情况下的做功课顺序。

如果多种顺序的最少扣分数都是最少的，那么输出字典序最小的那个。

最多只有15门功课，可以用状压DP。1表示完成，0表示未完成。

枚举所有情况 i ，如果一门功课 j 在 i 状态下已经完成了，那么我们比较状态 i - j 的扣分数和 i 的扣分数。

状态转移方程：

pass=dp[last].time+hw[j].cost-hw[j].limit;

dp[i].score=min(pass+dp[last].score , dp[i].score );

其中：

pass是在 i - j 状态下完成 j 功课的扣分数。（只是 j 功课的扣分数）

dp[ i ].score是 i 状态下的总最小扣分数。

dp[ i ].time是 i 状态下所用的总时间。

hw[ j ].cost是完成功课 j 所需的时间。

hw[ j ].limit是完成功课 j 的最晚时间。

下面贴代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<stack>using namespace std;#define INF 0x3f3f3f3f#define MAX 1<<15struct DP{    int pre,score,time,now;} dp[MAX];struct HW{    int cost,limit;    char name[111];}hw[33];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        memset(dp,0,sizeof(dp));        for(int i=0; i<n; i++)        {            scanf("%s%d%d",hw[i].name,&hw[i].limit,&hw[i].cost);        }        int End=1<<n;        for(int i=1;i<End;i++)        {            dp[i].score=INF;            for(int j=n-1;j>=0;j--)            {                //printf("ok\n");                int k=1<<j;                if(k&i)                {                    int last=i-k;                    int pass=dp[last].time+hw[j].cost-hw[j].limit;                    if(pass<0)                    {                        pass=0;                    }                    if(pass+dp[last].score<dp[i].score)                    {                        dp[i].score=pass+dp[last].score;                        dp[i].now=j;                        dp[i].pre=last;                        dp[i].time=dp[last].time+hw[j].cost;                    }                }            }        }        stack <int> s;        int tmp=End-1;        printf("%d\n",dp[tmp].score);        while(tmp)        {            s.push(dp[tmp].now);            tmp=dp[tmp].pre;        }        while(!s.empty())        {            printf("%s\n",hw[s.top()].name);            s.pop();        }    }    return 0;}