ACM HDU 1074 Doing Homework(位运算，搜索，状态压缩DP）

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1950    Accepted Submission(s): 685

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

 

 

算法核心：状态压缩DP
大意：
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：
n<=15，由题意知，只需对这n份作业进行全排列，选出扣分最少的即可。
用一个二进制数存储这n份作业的完成情况，第1.。。。n个作业状况分别
对应二进制数的第0，1.。。。。,n-1位则由题意，故数字上限为2^n
其中 2^n-1即为n项作业全部完成，0为没有作业完成。。。

用dp[i]记录完成作业状态为i时的信息（所需时间，前一个状态，最少损失的分数）。
递推条件如下
1.状态a能做第i号作业的条件是a中作业i尚未完成，即a&i=0。
2.若有两个状态dp[a],dp[b]都能到达dp[i],那么选择能使到达i扣分小的那一条路径，若分数相同，转入3
3.这两种状态扣的分数相同，那么选择字典序小的，由于作业按字典序输入，故即dp[i].pre = min(a,b);

初始化：dp[0].cost = 0;dp[0].pre=-1;dp[0].reduced = 0;

最后dp[2^n-1].reduced即为最少扣分，课程安排可递归的输出

/*
HDU1074
*/
#include<stdio.h>
#include<string.h>
const int MAXN=1<<16;
struct Node
{
    int cost;//所需时间
    int pre;//前一状态
    int reduced;//最少损失的分数 
}dp[MAXN];
bool visited[MAXN];
struct Course
{
    int deadtime;//截止日期
    int cost;//所需日期
    char name[201]; 
}course[16];
void output(int status)
{
    int curjob=dp[status].pre^status;
    int curid=0;
    curjob>>=1;
    while(curjob)
    {
        curid++;
        curjob>>=1;
    }
    if(dp[status].pre!=0)
    {
        output(dp[status].pre);
    } 
    printf("%s\n",course[curid].name);       
}    
int main()
{
    int T,n,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int upper=1<<n;
        int dayupper=0;
        for(i=0;i<n;i++)
        {
            scanf("%s%d%d",&course[i].name,&course[i].deadtime,&course[i].cost);
            dayupper+=course[i].cost;
        }  
        memset(visited,false,sizeof(visited));
        dp[0].cost=0;
        dp[0].pre=-1;
        dp[0].reduced=0;//dp[0]是指所有作业都没有做的状态  
        visited[0]=true;
        int work;
        int tupper=upper-1;//tupper表示成二进制数是n个1的，表示所有的作业都完成了 
        for(j=0;j<tupper;j++)//遍历所有状态 
        {
            for(work=0;work<n;work++)
            {
                int cur=1<<work;
                if((cur&j)==0)//该项作业尚未做过
                {
                    int curtemp=cur|j;
                    int day=dp[j].cost+course[work].cost;
                    dp[curtemp].cost=day;
                    int reduce=day-course[work].deadtime;
                    if(reduce<0)reduce=0;
                    reduce+=dp[j].reduced;
                    if(visited[curtemp])//该状态已有访问信息
                    {
                        if(reduce<dp[curtemp].reduced)
                        {
                            dp[curtemp].reduced=reduce;
                            dp[curtemp].pre=j;
                        } 
                      //else if(reduce==dp[curtemp].reduced)
                      //扣分相同，取字典序小的那一个，由于这里j是按从小到达搜索的，默认已是按字典序，不需再处理         
                        // {                                      
                            //  if(dp[curtemp].pre>j)        
                            //   dp[curtemp].pre = j;        
                            // }   
                    }
                    else //没有访问过
                    {
                        visited[curtemp]=true;
                        dp[curtemp].reduced=reduce;
                        dp[curtemp].pre=j;
                    }         
                }     
            }    
        }
        printf("%d\n",dp[tupper].reduced);   
        output(tupper);//递归输出 
    }    
}