Doing Homework again （贪心）

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11332    Accepted Submission(s): 6661

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

333 3 310 5 131 3 16 2 371 4 6 4 2 4 33 2 1 7 6 5 4

 

Sample Output

035

 按分数排序，，

//（思路错了） // 按分数高->低 排序，记录当天之前是否用过 #include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct Node{int d,p;}w[1010]; bool cmp(Node x,Node y){if(x.p==y.p)return x.d<y.d;return x.p>y.p;}int main(){int t,n,visit[1010];scanf("%d",&t);while(t--){memset(visit,0,sizeof(visit));scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d",&w[i].d);for(int i=0;i<n;i++)scanf("%d",&w[i].p);sort(w,w+n,cmp);int sum=0;for(int i=0;i<n;i++){int tt=w[i].d; while(tt) { if(!visit[tt]) { visit[tt]=1; break; } //这天没用过，  tt--;//当天被用过，往前推移 } if(tt==0)//这天之前都被用过，完不成作业，被罚 sum+=w[i].p;  }printf("%d\n",sum);}return 0;}