HDU 1074 Doing Homework（状态压缩DP）

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6845    Accepted Submission(s): 2969

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

【思路】：n的最大值15，而2的15次方只有3W多，所以将所有的情况就能列举出所有的情况

这种思想就是状态压缩的基本思想

【困惑】：有个地方没明白，kuangbin在j的那层循环是用顺序的，这样也行， 顺序和逆序的区别，他用的是位运算以及一些标记数组之类的，以后再看看

/*分析：对于n种家庭作业，全部做完有n!种做的顺序 
但是n!太大了，且对于完成作业1,2,3和1,3,2和2,1,3和3,2,1和3,1,2来说 
完成它们消耗的天数一定是一样的，只是完成的顺序不同从而扣的分不同 
所以可以将完成相同的作业的所有状态压缩成一种状态并记录扣的最少分即可 
即：状态压缩dp 
对于到达状态i,从何种状态到达i呢？只需要枚举所有的作业 
假如对于作业k,i中含有作业k已完成，那么i可以由和i状态相同的状态仅仅是k未完成的 
状态j=i-(1<<k)来完成k到达,并且j一定比i小，如果状态从0枚举到2^n-1那么j一定是在i之前已经计算过的  
*/  
#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<cstdlib>
#include<algorithm>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#include<vector>
#define F first
#define S second
#define PI acos(-1.0)
#define E  exp(1.0)
#define INF 0xFFFFFFF
#define MAX -INF
#define len(a) (__int64)strlen(a)
#define mem0(a) (memset(a,0,sizeof(a)))
#define mem1(a) (memset(a,-1,sizeof(a)))
using namespace std;
template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
template<class T> inline T Min(T a, T b) { return a < b ? a : b; }
template<class T> inline T Max(T a, T b) { return a > b ? a : b; }
struct pp
{
	char name[110];
	int dead,cost,pre;
};
pp a[50];
struct kode
{
    int time,score,pre,now;
} dp[1<<15];
int main() {
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
    int n,T;
    scanf("%d",&T);
    while(T--)
    {
    	scanf("%d",&n);
    	for(int i=0;i<n;i++)
    	{
    		scanf("%s%d%d",a[i].name,&a[i].dead,&a[i].cost);
    	}
    	int t=1<<n;
    	mem0(dp);
    	for(int i=1;i<t;i++)//枚举到达状态i  
    	{
    		dp[i].score=99999999;//初始化到达状态i的扣分   
    		for(int j=n-1;j>=0;j--)//由于输入时按字符大小输入，而每次完成j相当于把j放在后面完成且下面判断是dp[i]>dp[i-w]+score
    		{
    			int w=1<<j; //所以是n-1开始，如果下面判断是dp[i]>=dp[i-w]+score则从0开始   
    			if(w&i)//状态i不存在作业j完成则不能通过完成作业j到达状态i  
    			{
    				int past=i-w;//i-w表示没有完成j的那个状态  
    				int st=dp[past].time+a[j].cost-a[j].dead;
    				if(st<0)st=0;//完成j被扣分数最小是0   
    				if(st+dp[past].score<dp[i].score)
    				{
    					dp[i].score=st+dp[past].score;
    					dp[i].pre=past;//到达状态i的前驱,为了最后输出完成作业的顺序 
    					dp[i].now=j;
    					dp[i].time=dp[past].time+a[j].cost;//到达状态i花费的时间  
    				}
    			}
    		}
    	}
    	stack<int>s;
    	int temp=t-1;
    	cout<<dp[temp].score<<endl;
    	while(temp)
    	{
    		s.push(dp[temp].now);
    		temp=dp[temp].pre;
    	}
    	while(!s.empty())
    	{
    		cout<<a[s.top()].name<<endl;
    		s.pop();
    	}
    }
}