HDU 1074 - Doing Homework - [状压DP]

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可以说是第一道状压DP题，至少感觉这道题目还是挺简单的，不算很难理解；

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1074

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

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.

 

题意：

有n门课，每门课有截止时间（deadline）和完成所需的时间，如果超过规定时间完成，每超过一天就会扣1分，问怎样安排做作业的顺序才能使得所扣的分最小；

 

题解：

首先，dp思想：

假设dp[ _ , _ , _ , …… , _ , _ , _ ]记录了：n门课，在一些完成了，一些未完成的状态下，当前已经花去的时间（time），以及，当前已经被扣掉的最少分数（rs）；

例如：有3门课，那么dp[0,0,0]代表了三门课作业还未做完的状态，dp[0,1,1]代表第二门和第三门的作业已经做完，第一门课作业还没做完的状态；

那么，我们状态转移方程：

　　

　　也就是说，dp[…1…]表示当前状态下第k门课已完成；

　　那么，在dp[…0…]状态下，“已花费的时间 + 完成第k门课所需的时间 - 这门课的deadline”表示当前状态下开始做第k门课的作业，会导致扣多少分（记为red_sco）。

　　如果red_sco + dp[…0…].rs 比 dp[…1…].rs 小，说明我们可以更新一下dp[…1…]，它的rs可以更小；

 

然后，由于一门课的作业，只有做完和还没做完两个状态，所以我们dp[ _ , _ , _ , …… , _ , _ , _ ]里的n个数字，合起来看就是一个二进制数；

那么我们把它转换成十进制，其实并不影响实际的状态转移操作，故我们可以对这个状态进行压缩（原本n位的二进制数字可以压缩的很小位数的十进制数），即状压DP；

把状压DP讲的如此的清晰，简直泼妇爱科特，忍不住自恋一把……

 

AC代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<stack>
 4 #define INF 0x3f3f3f3f
 5 using namespace std;
 6 int n;//n门课
 7 struct Course{
 8     string name;
 9     int cost,deadline;
10 }course[16];
11 
12 struct DP{
13     int time,rs;
14     int now,pre;//用以记录做作业的顺序
15 }dp[1<<15];
16 
17 int main()
18 {
19     int t;
20     scanf("%d",&t);
21     while(t--)
22     {
23         scanf("%d",&n);
24         for(int i=0;i<n;i++) cin>>course[i].name>>course[i].deadline>>course[i].cost;
25 
26         int ed_state=(1<<n)-1;//所有作业都做完的状态
27 
28         dp[0].time=0, dp[0].rs=0;
29         for(int state=1;state<=ed_state;state++)//遍历状态
30         {
31             dp[state].rs=INF;
32             for(int co=n-1;co>=0;co--)//从后往前遍历课程，可以保证在相同扣分的情况下字典序完成课程作业
33             {
34                 int k=1<<(n-1-co);
35                 if(state & k)//如果当前状态，这门课作业已完成
36                 {
37                     int pre_state=state-k;
38 
39                     int tmp=dp[pre_state].time+course[co].cost-course[co].deadline;
40                     if(tmp<0) tmp=0;
41                         //算出要扣的分
42                     if(tmp+dp[pre_state].rs < dp[state].rs)
43                     {
44                         dp[state].rs=tmp+dp[pre_state].rs;
45                         dp[state].time=dp[pre_state].time+course[co].cost;
46 
47                         dp[state].now=co;
48                         dp[state].pre=pre_state;
49                     }
50                 }
51             }
52         }
53         printf("%d\\n",dp[ed_state].rs);
54         stack<int> output;
55         for(int i=ed_state;i!=0;i=dp[i].pre) output.push(dp[i].now);
56         while(!output.empty())
57         {
58             cout<<course[output.top()].name<<endl;
59             output.pop();
60         }
61     }
62 }

 

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