HDU 1789 Doing Homework again【贪心】

题目来戳呀

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

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

题意：
每个作业都能一天做完，老师给定期限，完不成要扣分，求扣的最少分数。
思路：
先把分数由大到小排序，再按期限从小到大排序。从大分数的期限依次往前找，哪天有空就放到哪天做（book标记）。
可以手动写一下怎样分配的过程，光想是很晕的（摊手）。

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int book[1010];typedef struct node{    int deadline;    int score;}ss;int cmp(ss x,ss y){    if(x.score!=y.score)        return x.score>y.score;    else        return x.deadline<y.deadline;}int main(){    int n,i;    scanf("%d",&n);    while(n--)    {        int t;        ss a[1010];        scanf("%d",&t);        for(i=0;i<t;++i)            scanf("%d",&a[i].deadline);        for(i=0;i<t;++i)            scanf("%d",&a[i].score);        sort(a,a+t,cmp);        int sum=0;        int j;        memset(book,0,sizeof(book));        for(i=0;i<t;++i)        {            for(j=a[i].deadline;j>0;--j)//大分数的那一天往前找            {                if(!book[j])                {                    book[j]=1;                    break;                }            }            if(j==0)                sum+=a[i].score;        }        printf("%d\n",sum);    }    return 0;}

ps：题目理解的很快，却不会敲，结果老师上课走神的时候想通了2333333