HDU 1789 Doing Homework again

Doing Homework again

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13845    Accepted Submission(s): 8034

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

 

Author

lcy

 

Source

2007省赛集训队练习赛（10）_以此感谢DOOMIII

 

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写一个结构体，每个单项包括截止时间、罚分

先对罚分进行降序排序，再对罚分相同时的截止时间进行降序排序。

从每个单项的截止时间开始向前搜索是否有空闲时间，有的占用其空闲时间，没有则加入总罚分

因为是降序排列，所以每次安排时间都先安排罚分重的，保证了结果最优

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;struct point{int x,y;};int cmp(const point &a,const point &b){if(a.y!=b.y)return a.y>b.y;return a.x>b.x;}int vis[1010];point a[1010];int main(){int t,sum,n,i,j,flag;scanf("%d",&t);while(t--){sum=0;memset(a,0,sizeof(a));memset(vis,0,sizeof(vis));scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i].x);}for(i=0;i<n;i++){scanf("%d",&a[i].y);}sort(a,a+n,cmp);for(i=0;i<n;i++){flag=0;for(j=a[i].x;j>0;j--){if(!vis[j]){vis[j]=1;flag=1;break;}}if(!flag) sum+=a[i].y;}printf("%d\n",sum);}return 0;}