hdu1789 Doing Homework again （贪心+优先队列）

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 9133    Accepted Submission(s): 5406

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

333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4

 

Sample Output

035

 

Author

lcy

 

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

 

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解析：按照期限从小到大、分数从大到小进行排序，然后依次扫描，如果p[i].date>=s(当前已处理作业数)，如果前面第 j 天的分数最小，并且该分数<p[i].score，那么就把 j 天的作业不做，腾出来做p[i]；否则，p[i]的作业不做。如果p[i].date>s，那么p[i]的作业要做。

          可以用优先队列存储已做作业的分数，然后进行处理。

代码：

#include<cstdio>#include<cstring>#include<cctype>#include<algorithm>#include<queue>using namespace std;const int maxn=1e3;struct tnode{int date,score;}p[maxn+10];int getin(){  int ans=0;char tmp;  while(!isdigit(tmp=getchar()));  do ans=(ans<<3)+(ans<<1)+tmp-'0';  while(isdigit(tmp=getchar()));  return ans;}bool cmp_p(tnode a,tnode b){  if(a.date!=b.date)return a.date<b.date;  return a.score>b.score;}int main(){  freopen("1.in","r",stdin);  int t,n,i,s,ans;  for(t=getin();t;t--)    {      n=getin();      for(i=0;i<n;i++)p[i].date=getin();      for(i=0;i<n;i++)p[i].score=getin();      sort(p,p+n,cmp_p);      priority_queue<int, vector<int>, greater<int> > q;      for(ans=s=0,i=0;i<n;i++)        {  if(p[i].date>s)    {      q.push(p[i].score);      s++;  continue;}          if(q.top()>=p[i].score)ans+=p[i].score;  else ans+=q.top(),q.pop(),q.push(p[i].score);       }  printf("%d\n",ans);}  return 0;}