Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10043 Accepted Submission(s): 5875
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.
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
好吧这题我只知道要根据分数排序，写完bool cmp后sort就不知道咋办了。弱比的我还是得百度。
做法：首先根据分数降序排序，其次若分数相同，则天数少的排前面（升序）。然后再创建一个记录int数组，由于是贪心问题，尽量将要求第几天完成的排在那一天，给前面腾出空间若那一天被占用了，则往前找空的一天，原则就是尽量往后排，尽量拖延，能迟点完成就迟点完成（尽量往后排）
代码：
#include<iostream>
#include<algorithm>
using namespace std;
struct id
{
int day;
int fenshu;
};
bool cmp(const id a,const id b)
{
if(a.fenshu!=b.fenshu)
return a.fenshu>b.fenshu;//先降序排分数
else return a.day<b.day;//再升序排截止日期
}
int main(void)
{
int n,t,i,use[1009],j,sum,tem;
cin>>t;
while(t--)
{
memset(use,0,sizeof(use));
cin>>n;
id *haha=new id[n+2];//new一下节省空间,n+2有安全感....
for (i=0; i<n; i++)
cin>>haha[i].day;
for (i=0; i<n; i++)
cin>>haha[i].fenshu;
sort(haha,haha+n,cmp);//排个序
sum=0;
for (i=0; i<n; i++)
{
tem=haha[i].day;//这里比较重要，从某分数对应的天数开始往前找
while(tem)
{
if(use[tem]==0)//为0则没被占用
{
use[tem]=1;//那现在就要占用了，标记为1
break;//果断break
}
else
{
tem--;//自己对应那天被占用则往前找（占用别人的）
}
}
if(tem==0)//到0了还没找到，那就注定没时间补了
sum+=haha[i].fenshu;//扣下这分！
}
cout<<sum<<endl;
delete []haha;//记得delete[]
}
return 0;
}