HDU-1789

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.
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 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：ACM大佬比赛回来但无奈作业都没写完，现在给你他的每门课程的deadline和如果不写完所扣的分数。现在要求你这样安排才能使扣分最少。输出最少的分数。
思路：贪心算法。先把分数按从大到小排序，然后将最大的放在它的最后的期限做。依次来如果deadline有作业可做，就往前进一个直到第一天为止。
AC代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;struct homework{    int dl;    int score;    bool flge;} hw[1005];bool vis[1005];bool cmp(homework a,homework b){    return a.score>b.score;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%d",&hw[i].dl);        }        for(int i=0;i<n;i++)        {            scanf("%d",&hw[i].score);        }        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            hw[i].flge=0;        }        sort(hw,hw+n,cmp);        for(int i=0;i<n;i++)        {            for(int j=hw[i].dl;j>0;j--)            {                if(vis[j]==0)                {                    vis[j]=1;                    hw[i].flge=1;                    break;                }            }        }        int sum=0;        for(int i=0;i<n;i++)        {            if(hw[i].flge==0)            {                sum+=hw[i].score;            }        }        printf("%d\n",sum);    }    return 0;}