Doing Homework again——贪心

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

 

思路

先按照正常思维排序，依次完成任务，到后面不能够完成的任务，要考虑是否可以替换掉前面已完成的任务从而减少惩罚。既然天数一定不够用，既然一定要惩罚一门，那么就替换一门惩罚更大的出来。

代码

#include
using namespace std;

struct node{
	int dl,rd,vis;
}a[1005];

bool cmp(node a,node b){
	if(a.dl==b.dl) return a.rd>b.rd;
	return a.dl>t;
	while(t--){
		cin>>n;
		for(int i=0;i>a[i].dl;
		}
		for(int i=0;i>a[i].rd;
			a[i].vis=0;
		}
		sort(a,a+n,cmp);
		int day=1,sum=0;
		
		for(int i=0;i=day){
				a[i].vis=1;
				day++;
				continue;
			}
			int minn=a[i].rd,index=i;
			for(int j=0;ja[j].rd){   //在已完成中选一个最小的替换
					minn=a[j].rd;
					index=j;
				}
			}
			if(i!=index){
				a[i].vis=1;
				sum+=a[index].rd;
				a[index].vis=0;    //记得恢复状态
			}else{
				sum+=a[i].rd;   
			}
		}
		cout << sum <