HDU1074-Doing Homework

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#状压dp

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本文介绍了一道经典的状压动态规划问题，即如何合理安排做作业的顺序以最小化因延迟提交而被扣分的情况。文章通过一个具体的示例详细解析了问题背景、输入输出要求、解题思路及实现代码。

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Doing Homework

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9938 Accepted Submission(s): 4764

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

题目大意：有门科目，每门科目有一个截止时间和完成花费时间，每超出截止时间一天，分数减少一分，问在完成所有科目的前提下，如何安排科目顺序使得分数减少最少，按字典序输出科目顺序。
解题思路： 状压dp，因为只有15门课，所以将科目的完成情况记做一个状态，然后枚举科目进行转移。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=(1<<15)+5;
const int MOD=1e9+7;
int dead[20],cost[20];
int dp[MAXN],t[MAXN];
char s[20][105];
int pre[MAXN];

void print(int x)
{
  if(x==0) return;
  print(x^(1<<pre[x]));
  printf("%s\n",s[pre[x]] );
}

int main()
{
  int T;
  scanf("%d",&T );
  while(T--)
  {
    memset(pre,0,sizeof(pre));
    int n;
    scanf("%d",&n );
    for(int i=0;i<n;i++)
    {
      scanf("%s%d%d",s[i],&dead[i],&cost[i] );
    }
    // for(int i=0;i<n;i++)
    // cout<<s[i]<<endl;
    for(int i=1;i<(1<<n);i++)
    {
      dp[i]=INF;
      for(int j=n-1;j>=0;j--)
      {
        int tmp=1<<j;
        if(!(i&tmp)) continue;
        int score=t[i^tmp]+cost[j]-dead[j];
        if(score<0) score=0;
        if(dp[i]>dp[i^tmp]+score)
        {
          dp[i]=dp[i^tmp]+score;
          t[i]=t[i^tmp]+cost[j];
          pre[i]=j;
        }
      }
    }
    printf("%d\n", dp[(1<<n)-1]);
    print((1<<n)-1);
  }
  return 0;
}