刷题总结——doing homework again（hdu1789）

题目：

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

题解：

贪心····直接按扣的分数从大到小排序··如果分数相等就按截止日期从小到大排序····然后从第一项作业开始枚举··每此从这项作业的截止日期往以前的天枚举··第一次枚举到的空闲天数作为这项作业的完成天数··然后打上标记即可··如果找不到空闲天数ans加上这项作业的扣分

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
using namespace std;
const int N=;
struct node
{
  int dl,re;
}hw[N];
int T,n;
inline int R()
{
  char c;int f=;
  for(c=getchar();c<''||c>'';c=getchar());
  for(;c<=''&&c>='';c=getchar())  f=(f<<)+(f<<)+c-'';
  return f;
}
inline bool cmp(node a,node b)
{
  if(a.re==b.re)  return a.dl<b.dl;
  return a.re>b.re;
}
bool visit[];
int main()
{
 // freopen("a.in","r",stdin);
  T=R();
  while(T--)
  {
    int ans=;
    memset(visit,false,sizeof(visit));
    n=R();
    for(int i=;i<=n;i++) hw[i].dl=R();
    for(int i=;i<=n;i++) hw[i].re=R();
    sort(hw+,hw+n+,cmp);
    for(int i=;i<=n;i++)
    {
      bool flag=false;
      for(int j=hw[i].dl;j>=;j--)
        if(!visit[j])
        {
          visit[j]=true,flag=true;break;
        }
      if(flag)  continue;
      else ans+=hw[i].re;
    }
    cout<<ans<<endl;
  }
  return ;
}