HDOJ-1789 Doing Homework again

Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11329 Accepted Submission(s): 6658

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

此题是一个比较复杂的贪心问题，先将分数从大到小排序，分数一样以期限从小到大排，然后按照开始日期从后往前找没有占用的日期，如果找不到了，那么记录这个分数，并把它加在最后扣除的分数里面。

代码如下：

#include <cstdio>#include <algorithm>#include<stdint.h>#include<math.h>#include<string.h>struct s{    int d,x;}p[1010];int a[10000];bool cmp(s q,s r){    if(q.x!=r.x)return q.x>r.x;    return q.d<r.d;}using namespace std;int main(){    int t,n,i,j,sum;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        memset(p,0,sizeof(p));        memset(a,0,sizeof(a));        sum=0;        for(i=0;i<n;i++) scanf("%d",&p[i].d);        for(i=0;i<n;i++) scanf("%d",&p[i].x);        sort(p,p+n,cmp);        for(i=0;i<n;i++){            for(j=p[i].d;j>0;j--){                if(!a[j]){                    a[j]=1;                    break;                }            }            if(!j) sum+=p[i].x;        }        printf("%d\n",sum);    }    return 0;}