HDOJ1074 Doing Homework[DP+状态压缩]

Doing Homework

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3150    Accepted Submission(s): 1203

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

 

 

 

 

 

 

 

 

 

 

第一次测试数据：

代码是看了一位大牛的，和OY一起慢慢调试总算搞清楚，还是很有手滑的，画了个图，便于理解。关于DP和状态压缩还是需要多练练，以前都没接触过！状态压缩很巧妙！

思路：说说第一个测试数据，总共有3门课，也就是总共有7状态，而3，5，6状态是建立在状态1，2，4上的，通过自下而上进行DP，保留每个状态中的四个记录，一直到最后一个状态。

code：

#include <iostream> 
#include <cstdio> 
#include <cstdlib> 
#include <ctime> 
#include <map> 
#include <cstring> 
#include <string.h> 
#include <fstream> 
#include <vector> 
#include <cmath> 
#include <algorithm> 
  
using namespace std; 

const int INF = 1<<30; 
const int N = 20; 
  
struct work{ 
    char name[N];    //课程名字 
    int deadline;   //截止时间 
    int use_time;   //需要时间 
}; 
  
work data[N+1]; 
  
struct state{ 
    int used_time;      //已经使用时间 
    int reduce_score;   //减掉的最小学分 
    int last_work;      //此状态下前一个工作 
    int ith_work;       //这个状态完成的最后一个工作 
}; 
  
state d[1<<N]; 
  
int main()
{   
    int tot, n, i, j, max_value; 
    scanf("%d", &tot); 
    while (tot--) 
    { 
        scanf("%d", &n); 
        for (i=0; i<n; i++)
        { 
            scanf("%s",&data[i].name); 
            scanf("%d%d", &data[i].deadline, &data[i].use_time); 
        }   
        max_value = (int)(1<<n);
        int temp, score, last, reduce;
        d[0].reduce_score = 0;
        d[0].used_time = 0;
        d[0].last_work = -1;
        for (i=1; i<max_value; i++)                                                //相当于自下而上进行DP
        {  
            d[i].reduce_score = INF; 
            /* 
                为什么j要从后往前枚举： 
                比如i的二进制=11，则应考虑1、完成01后最小花费+10的花费 
                                            2、完成10后最小花费+01的花费 
                两个值较小的那一个，如果相等当然是取1，其他情况类推 
                这和题目要求的字符串升序排列有关，与最小值无关 
            */
            for (j=n-1; j>=0; j--)
            {
                temp = (int)(1<<j);
                //如果状态i中完成了第j个作业
                if (i & temp)
                { 
                    //如果i==temp即只完成j则last=0；
                    //则d[0].reduce_score = 0;
                    //  d[0].used_time = 0; 
                    //  d[0].last_work = -1; 
                    //如果i!=temp,则
                    //比如i=7，设被扣学分score[7]=min(score[3]+完成3后完成4被扣的学分，
                    //                                score[6]+完成6后完成1被扣的学分，
                    //                                score[5]+完成5后完成2被扣掉的学分)
                    last = i-temp;
                    //如果有足够的时间达到j状态，则求得的是在这个状态完成并被扣学分最小的情况下完成这个作业被扣的学分
                    //                           否则则置为0
                    //      上一个状态所花的时间 + 完成第j个课程所花的时间 - 这门课的最后期限
                    reduce = d[last].used_time   +  data[j].use_time       - data[j].deadline; 
                    reduce = reduce>0?reduce:0;
                     //(没有完成这个作业的某个状态被扣的学分 + 在这个状态完成并被扣学分最小的情况下完成这个作业被扣的学分) < i状态下被扣学分，就更新
                    if (d[last].reduce_score                 + reduce                                                      < d[i].reduce_score) 
                    {
                        d[i].reduce_score = d[last].reduce_score+reduce;
                        d[i].last_work = last;
                        d[i].ith_work = j;
                        d[i].used_time = d[last].used_time+data[j].use_time;
                    }
                }
            }
        }
        printf("%d\n", d[max_value-1].reduce_score); 
        int pos[N+1], k=0, x=max_value-1; 
        //递推得出序列 
        while (x > 0)
        { 
            pos[k++] = d[x].ith_work; 
            x = d[x].last_work; 
        }  
        for (i=k-1; i>=0; i--) 
            printf("%s\n",data[pos[i]].name); 
    } 
    return 0; 
}