Doing Homework HDU - 1074 (状压DP)

本文介绍了一个通过状态压缩动态规划解决的作业排序问题,旨在帮助学生在有限时间内完成多门课程作业,减少因延期提交而被扣分的情况。文章提供了一段完整的C++代码实现,展示了如何有效地安排作业完成顺序。

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Doing Homework

                                                                                                        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
23Computer 3 3English 20 1Math 3 23Computer 3 3English 6 3Math 6 3
 

Sample Output
2ComputerMathEnglish3ComputerEnglishMath
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
 
1. 首先状压DP涉及位运算, 先补一下位运算;

    a&b 、a|b  :  a和b写成二进制每一位做与运算,再转成十进制;

dp[ i ]表示状态 i 下有那几个是最优的;

pre[j]记录状态 j 的前一个状态

代码:

 

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f  //1061109567
#define ll long long
using namespace std;
const int MAXN=16;
struct Node
{
    char name[110];
    int D,C;
}node[MAXN];

int dp[1<<MAXN];
int pre[1<<MAXN];
int n;

void output(int status)
{
    if(status==0)return;
    int t=0;
    for(int i=0;i<n;i++)
      if((status&(1<<i))!=0 && (pre[status]&(1<<i))==0)
      {
          t=i;
          break;
      }
    output(pre[status]);
    printf("%s\n",node[t].name);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 0;i < n; i++) scanf("%s%d%d",&node[i].name,&node[i].D,&node[i].C);
        memset(dp, INF, sizeof(dp));
        dp[0]=0;
        for(int i=0;i<(1<<n);i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i&(1<<j))continue;
                int s=0;
                for(int k=0;k<n;k++)
                  if(i&(1<<k))
                     s+=node[k].C;
                s+=node[j].C;
                if(s>node[j].D)s=s-node[j].D;
                else s=0;
                if(dp[i|(1<<j)]>dp[i]+s)
                {
                    dp[i|(1<<j)]=dp[i]+s;
                    pre[i|(1<<j)]=i;
                }
            }
        }
        printf("%d\n",dp[(1<<n)-1]);
        output((1<<n)-1);
    }
    return 0;
}

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