HDU 1789 Doing Homework again (貪心)

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Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6499    Accepted Submission(s): 3874

Problem DescriptionIgnatius 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. 

 

InputThe 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. 

 

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

 

Sample Input333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4 

 

Sample Output035 

 

Authorlcy 

 

Source2007省賽集訓隊練習賽（10）_以此感謝DOOMIII 

 

Recommendlcy  這道題的貪心策略是先取減分多的，然後取減分少的所以先要排序，把減分多的排到前面，如果減分相同，把deadline小的放前面。然後我每次取出一個任務，如果能在deadline那一天執行，就放在那一天執行，不行的話，就往前找，找到能執行的那一天就在那一天執行。如果所有能執行的天數都有其他的任務的話，那麼我這個任務就不能完成，就要減去相應的分數。

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 struct node 7 { 8     int day; 9     int num;10     bool operator<(const node &B)const11     {12         if(num!=B.num)13             return num>B.num;14         else15             return day<B.day;16     }17 }a[1000];18 int vis[1000];19 int main()20 {21     int kase;22     scanf("%d",&kase);23     while(kase--)24     {25         int n,i,j,sum=0,ans=0;26         memset(vis,0,sizeof(vis));27         scanf("%d",&n);28         for(i=1;i<=n;i++)29             scanf("%d",&a[i].day);30         for(i=1;i<=n;i++)31             scanf("%d",&a[i].num);32         sort(a+1,a+n+1);33         for(i=1;i<=n;i++)34         {35             for(j=a[i].day;j>0;j--)36             {37                 if(!vis[j])38                 {39                     vis[j]=1;40                     break;41                 }42             }43             if(j==0)44                 ans+=a[i].num;45         }46         printf("%d\n",ans);47     }48     return 0;49 }

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