HDU 1074 Doing Homework(狀壓DP)

Mr_Treeeee發表於2020-04-06

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9486    Accepted Submission(s): 4481


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



 




題意：


給出每個作業的截止日期和長度，晚了截止日期幾天完成就扣幾分。求最小扣分。


POINT：


Nmax=15，所以用狀態壓縮做。題目裡說了用字典序，但是它自己已經排好了，所以不用管。





#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <math.h>
using namespace std;
#define ll long long
struct node
{
    int day;
    int score;
    int pre;
}dp[1<<15];
struct ode
{
    string name;
    int end;
    int l;
}a[16];
int flag[1<<15];
void pree(int x)
{
    stack<int> q;
    while(x!=0)
    {
        q.push(x);
        x=dp[x].pre;
    }
    int now=0;
    while(!q.empty())
    {
        int w=now^q.top();
        int wei=log(w)/log(2);
        cout<<a[wei].name<<endl;
        now=q.top();
        q.pop();
    }
    
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        memset(dp,0,sizeof dp);
        memset(flag,0,sizeof flag);
        for(int i=0;i<n;i++)
        {
            cin>>a[i].name>>a[i].end>>a[i].l;
        }
        dp[0].day=0;
        dp[0].pre=-1;
        dp[0].score=0;
        for(int i=0;i<=(1<<n)-1;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(((1<<j)&i)==1) continue;
                int newi=(1<<j)|i;
                int newday=dp[i].day+a[j].l;
                int addscore=0;
                if(newday>a[j].end)
                {
                    addscore=newday-a[j].end;
                }
                
                int newscore=dp[i].score+addscore;
                if(flag[newi])
                {
                    if(newscore<dp[newi].score)
                    {
                        dp[newi].day=newday;
                        dp[newi].pre=i;
                        dp[newi].score=newscore;
                    }
                }
                else
                {
                    flag[newi]=1;
                    dp[newi].day=newday;
                    dp[newi].pre=i;
                    dp[newi].score=newscore;
                }
            }
        }
        printf("%d\n",dp[(1<<n)-1].score);
        pree((1<<n)-1);
    }
}








                                    




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