Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13847    Accepted Submission(s): 8036

Problem DescriptionIgnatius 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. InputThe 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. OutputFor each test case, you should output the smallest total reduced score, one line per test case. Sample Input333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4Sample Output0 35

/*    Name: hdu--1798--Doing Homework again    Copyright: 2017 日天大帝    Author: 日天大帝    Date: 21/04/17 15:32    Description: 贪心，思路让当前分数大的替换当前分数小的作业*/#include<iostream>#include<queue>#include<cstring>#include<algorithm>using namespace std;struct work{    int score,deadline;    bool operator<(const work &a)const{        return score>a.score;    }}arr[];bool cmp(work a,work b){    return a.deadline<b.deadline;}priority_queue<work> q;//按照分数排序int main(){    ios::sync_with_stdio(false);    int T;cin>>T;    while(T--){        memset(arr,,sizeof(arr));        while(!q.empty())q.pop();        int n;cin>>n;        ; i<n; ++i)cin>>arr[i].deadline;        ; i<n; ++i)cin>>arr[i].score;        ,ans = ;        sort(arr,arr+n,cmp);//按照时间排序        ; i<n; ++i){            q.push(arr[i]);            if(t < arr[i].deadline){                t++;continue;            }            ans += q.top().score;            q.pop();        }        cout<<ans<<endl;    }    ;}

/*大神的代码，优先队列和排序很巧妙*/#include <cstdio>#include <cstring>#include <iostream>#include <string>#include <algorithm>#include <map>#include <set>#include <queue>#include <utility>#include <vector>#include <iterator>using namespace std;typedef long long ll;typedef pair<int, int> P; << ;const int INF = 0x3f3f3f3f;P arr[MAX_N];int main() {    //ios::sync_with_stdio(false);    //cin.tie(NULL);    //cout.tie(NULL);    int T;    scanf("%d", &T);    while (T--) {        int n;        scanf("%d", &n);        ; i < n; ++i)            scanf("%d", &arr[i].first);        ; i < n; ++i)            scanf("%d", &arr[i].second);        sort(arr, arr + n);        , day = ;        priority_queue<int, vector<int>, greater<int> > pque;        ; i < n; ++i) {            pque.push(arr[i].second);            if (day < arr[i].first) {                ++day;                continue;            }            ans += pque.top();            pque.pop();        }        printf("%d\n", ans);    }    ;}