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.
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
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_pair
const 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 10007
typedef 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;
}