HDOJ 1789

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4668    Accepted Submission(s): 2737

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

 

Recommend

lcy

 

这题数据水了，1000的范围完全可以n^2的贪过去，数据范围要是大一点就不好做了。

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>#include <iomanip>using namespace std;//#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define lson l , mid  , rt << 1#define rson mid + 1 , r , rt << 1|1#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)#define mid ((l + r) >> 1)#define mk make_pairconst int MAXN = 100000 + 50;const int maxw = 100 + 20;const int MAXNNODE = 1000 +10;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x7fffffff;const int IMIN = 0x80000000;#define eps 1e-8#define mod 10007typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pii;const D e = 2.718281828459;int day[MAXN] , cost[MAXN];int n , sum , max_day;int main(){    //ios::sync_with_stdio(false);#ifdef Online_Judge    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif // Online_Judge    int T , key , ans;    cin >> T;    while(T--)    {        max_day =  ans = sum = 0;        scanf("%d" , &n);        FOR(i , 0 , n)        {            scanf("%d" , &day[i]);            max_day = max(max_day , day[i]);        }        FOR(i ,  0  , n)        {            scanf("%d" , &cost[i]);            sum += cost[i];        }        REP(i , max_day , 0)        {            int tmp = 0;            FOR(j , 0 , n)            if(day[j] >= i&& cost[j] > tmp)            {                tmp = cost[j];                key = j;            }            cost[key] = 0;            ans += tmp;        }        printf("%d\n" , sum - ans);    }    return 0;}