HDU1789时间贪心

Posted 2020-09-15 zzq

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13883    Accepted Submission(s): 8053

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

 想到了是贪心，但一直以为时间可能会给的很大不敢直接按时间一个一个遍历，没想到最后真的这样A的，恶心的题目为啥不说清数据范围！

时间段是不会变化的，所以我们要充分利用好每一个时间段，即在这个时间段里让其获得最大的利益，由于彼此天数之间毫无关系所以可以贪心每一天。

用一个数组记录这一天是否被使用，将结构体排序，排序方式:先按照分数大小排序，如果分数一样，天数大的排在前面；

贪心方法:找出当前某门科目，从这门课的截止日期开始往前推只要遇到没标记的天就占用！

#include<bits/stdc++.h>
using namespace std;
int vis[10005];
struct node
{
int a,b;
}P[1005];
bool cmp(node A,node B)
{
if(A.b==B.b) return A.a>B.a;
else return A.b>B.b;
}
int main()
{
int t,n,m,i,j;
cin>>t;
while(t--){memset(vis,0,sizeof(vis));
cin>>n;int ans=0,sumn=0;
for(i=1;i<=n;++i) cin>>P[i].a;
for(i=1;i<=n;++i) cin>>P[i].b,sumn+=P[i].b;
sort(P+1,P+1+n,cmp);
//for(i=1;i<=n;++i) cout<<P[i].a<<" "<<P[i].b<<endl;
for(i=1;i<=n;++i){
for(j=P[i].a;j>=1;j--){
if(!vis[j]) {vis[j]=1;ans+=P[i].b;break;}
}
}//cout<<ans<<endl;
cout<<sumn-ans<<endl;
}
return 0;
}