hdu1789经典的贪心

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13252    Accepted Submission(s): 7725

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

 

Author

lcy

 

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

按分数排序，分数一样就让期限短的在前，然后居然是一个一个枚举...用vis数组标记 j=deal[i]记录该门课的期限，然后往前走，如果能找到一个vis为0的 也就是那一天可以安排 如果前面几天全是1了 那么就没法完成了 大概这个思路，真难想，看了别人的思路再自己写的。。。我太水了

#include
#include
#include
using namespace std;
int deal[105];
int score[105];
int vis[105];
int main(){
	int t,i,j,m,n;
	cin>>t;
	while(t--){
		scanf("%d",&n);
		for(i=1;i<=n;i++){
			scanf("%d",&deal[i]);
		}
		for(i=1;i<=n;i++){
			scanf("%d",&score[i]);
		}
		for(i=1;i<=n;i++)
		for(j=i+1;j<=n;j++){
			if(score[i]deal[j]) swap(deal[i],deal[j]);
			}
		}
		memset(vis,0,sizeof(vis));
		int ans=0;
	    for(i=1;i<=n;i++){
	    	for(j=deal[i];j>=0;j--)
	    	{
				if(vis[j]==0){
					vis[j]=1;
					break;
				}
			}
				if(j<=0) ans+=score[i];
		}
       cout<