Hduoj1789 【贪心】

/*Doing Homework againTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6852    Accepted Submission(s): 4077Problem 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 4 Sample Output035 Authorlcy Source2007省赛集训队练习赛（10）_以此感谢DOOMIII */#include<stdio.h>#include<stdlib.h>#include<string.h>struct hm{int x, y;}f[1010];int cmp(const void*a, const void *b){return (*(struct hm*)b).y - (*(struct hm*)a).y;}int main(){int i , j, k, m, n, ans[1010];scanf("%d", &n);while(n--){scanf("%d", &m);for(i = 1; i <= m; i++)scanf("%d", &f[i].x);for(i = 1; i <= m; i++)scanf("%d", &f[i].y);qsort(f+1, m, sizeof(f[0]), cmp);memset(ans, 0, sizeof(ans));k = 0;for(i = 1; i <= m; i++){for(j = f[i].x; j >= 1; j--){if(!ans[j]){ans[j] = 1;break;}}if(j < 1)k += f[i].y;}printf("%d\n", k);}return 0;} 

难点：此题的关键在于找出最优序列该怎么排，主要是按照分数的大小来，其次关键是完成这项作业的最优时间就是期限当天，若是无法在期限当天完成，则将时间往前推移，直至没时间去完成这项作业，则将此项作业的分数加到损失里面去。