HDU1789 解题报告

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14505    Accepted Submission(s): 8452

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

 

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解法：贪心

将所有的作业按照ddl从大到小排列

从最大的ddl开始，天数依次递减，直到第一天为止

举个例子

假设当前的ddl为5，则将ddl大于或者等于5的，而且没做过的作业中reduce最大的取出来做掉

然后将ddl减小为4，重复上述步骤……

#include<iostream>#include<algorithm>using namespace std;struct homework{    int ddl;    int rdc;    bool used;};const int maxn=1000+10;homework hk[maxn];int select[maxn];int n;int rdc_all;bool cmp(const homework &hk1,const homework &hk2);void solve();int main(){    int t;    cin>>t;    while(t--)    {        cin>>n;        rdc_all=0;        for(int i=0;i<n;i++) cin>>hk[i].ddl;        for(int i=0;i<n;i++) cin>>hk[i].rdc;        for(int i=0;i<n;i++)        {            rdc_all+=hk[i].rdc;            hk[i].used=false;        }        solve();    }    return 0;}bool cmp(const homework &hk1,const homework &hk2){    if(hk1.ddl!=hk2.ddl) return hk1.ddl>hk2.ddl;    else return hk1.rdc>hk2.rdc;}void solve(){    sort(hk,hk+n,cmp);    int max_ddl=hk[0].ddl;    while(max_ddl)    {        int max_rdc=0;        int max_rdc_index;        for(int i=0;i<n;i++)        {            if(hk[i].ddl<max_ddl) break;            if(!hk[i].used&&hk[i].rdc>max_rdc)            {                max_rdc=hk[i].rdc;                max_rdc_index=i;            }        }        hk[max_rdc_index].used=true;        rdc_all-=max_rdc;        max_ddl--;    }    cout<<rdc_all<<endl;}