Hdu 1789 Doing Homework again

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9969    Accepted Submission(s): 5830

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

思路：很经典的贪心题目，不过一开始没有思路，不晓得该怎么贪心，参考网上的思路：按照score从高到低排序，相等按照time从低到高排序，

然后依次选取，并且判断在当前时间的time内有没有空白的地方可以使用，如果有的话就使用那一天，如果没有，说明时间不足，该时间失败，记录分数

AC代码如下：

#include <iostream>#include <cstring>#include <algorithm>using namespace std;const int maxn=1000+5;struct Node{    int time,score;}pp[maxn];bool cmp(Node a,Node b){    if(a.score==b.score) return a.time<b.time;    return a.score>b.score;}int use[maxn];int main(){    int t;    cin>>t;    while(t--){        int n;        cin>>n;        for(int i=0;i<n;i++) cin>>pp[i].time;        for(int i=0;i<n;i++) cin>>pp[i].score;        sort(pp,pp+n,cmp);        memset(use,0,sizeof(use));        int ans=0;        int j;        for(int i=0;i<n;i++){            for(j=pp[i].time;j>0;j--){                if(!use[j]){                    use[j]=1;break;                }            }            if(j==0) ans+=pp[i].score;        }        cout<<ans<<endl;    }    return 0;}