Doing Homework again

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/38705511

本文介绍了一个基于价值调整的01背包问题解决方案，用于优化学生在有限时间内完成多项作业的顺序，以最小化未完成作业的损失分数。通过将问题转化为求解最大价值和，采用排序和动态规划的方法实现了高效求解。

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Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6610 Accepted Submission(s): 3950

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. And now we assume that doing everyone homework always takes one day. 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 that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.

Output

For each test case, you should output the smallest total reduced score, one line per test case.

Sample Input

3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4

Sample Output

0
3
5

题意：完成一个作业需要一天，每个作业都有相应的权值和最后期限，问你没有完成的作业的最小权值是多少。

思路：我们不妨转换成n天之内，你能完成的作业的最大权值和是多少，这里用到一个价值改变的01背包，当时间大于期限的时候我们可以把他的价值看成是0。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{ int val,dead;
}work[1010];
bool cmp(node a,node b)
{ return a.dead<b.dead;
}
int dp[1010],sum,ans;
int main()
{ int t,n,i,j,k;
  scanf("%d",&t);
  while(t--)
  { scanf("%d",&n);
    sum=0;
    memset(dp,0,sizeof(dp));
    for(i=1;i<=n;i++)
     scanf("%d",&work[i].dead);
    for(i=1;i<=n;i++)
    { scanf("%d",&work[i].val);
      sum+=work[i].val;
    }
    sort(work+1,work+1+n,cmp);
    for(i=1;i<=n;i++)
    { for(j=n;j>work[i].dead;j--)
       dp[j]=max(dp[j],dp[j-1]);
      for(j=work[i].dead;j>=1;j--)
       dp[j]=max(dp[j],dp[j-1]+work[i].val);
    }
    ans=0;
    for(i=1;i<=n;i++)
     ans=max(ans,dp[i]);
    printf("%d\n",sum-ans);
  }
}