Doing Homework again
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
//看了一些別人的代碼，才開始真的地有了一些想法，，剛開始在排序的過程中打算按時間（從小到大）排序的，若時間相同則按分數從大到小排序，結果發現有點小麻煩，有點寫不下去的感覺後來就又換了，，感覺按分數排序（從大到小）較為簡單，若分數相同，則按時間從小到大排序
#include
#include
#include
using namespace std;
struct zy
{
int day;
int score;
};
bool compare(zy x,zy y)
{
if(x.score==y.score)
return x.dayy.score;//將分數按從大到小排序
}
int main()
{
int T;
cin>>T;
zy a[2000];
while(T--)
{
int n,i,j;
cin>>n;
int flag[2000]={0};//標志無任務
for(i=0;i>a[i].day;
for(i=0;i>a[i].score;
sort(a,a+n,compare);
int sum=0;
for(i=0;i=1;j--)
{
if(flag[j]==0)
{
flag[j]=1;//代表第j天有已經有任務了
break;
}
}
if(j==0)//代表任務未完成
sum+=a[i].score;
}
cout<
