HDU(1789)Doing Homework again(贪心,找出损失最少分)
来源:互联网 发布:分类目录源码帝国cms 编辑:程序博客网 时间:2024/06/14 04:03
Doing Homework again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11306 Accepted Submission(s): 6636
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.
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题目大意:给出n道题,并给出每道题完成的限制时间,以及不完成所损失的分数,每天只能完成一道,学生如何做才能使损失的分数最少。解法,用结构体把每题的限制时间以及损失的分数,然后按分数大小排序,在从最大分数的限制往前搜,并将用过的天数标记;#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;struct none{int x,y;}no[10000];bool cmp(none a, none b){return a.y>b.y;}int main (){int t,n;scanf("%d", &t);while(t--){int i, j, sum=0;int a[1001];memset(a,0,sizeof(a));int k=0;scanf("%d", &n);for(i=0; i<n; i++)scanf("%d", &no[i].x);for(i=0; i<n; i++)scanf("%d", &no[i].y);sort(no,no+n,cmp);for(i=0; i<n; i++){k=0;for(j=no[i].x; j>0; j--){if(a[j]==0){a[j]=1;k++;break;}}if(k==0)sum+=no[i].y;}printf("%d\n", sum);}return 0;}
0 0
- HDU(1789)Doing Homework again(贪心,找出损失最少分)
- hdu 1789 Doing Homework again (贪心)
- hdu 1789 Doing Homework again(贪心)
- HDU 1789 Doing Homework again (贪心)
- HDU 1789 Doing Homework again (贪心)
- HDU 1789 Doing Homework again(贪心~!)
- hdu 1789 Doing Homework again (贪心)
- HDU 1789 Doing Homework again(贪心)
- HDU 1789 Doing Homework again(贪心)
- HDU 1789 Doing Homework again(贪心)
- hdu 1789 Doing Homework again(贪心)
- HDU-1789-Doing Homework again(贪心)
- HDU 1789 Doing Homework again(贪心)
- hdu 1789 Doing Homework again(贪心)
- HDU 1789 Doing Homework again(贪心)
- HDU-1789-Doing Homework again(贪心)
- hdu 1789 Doing Homework again 贪心算法
- hdu 1789 Doing Homework again 贪心算法
- uva 11388 GCD LCM
- iOS原生二维码扫描&libqrencode生成二维码
- Android使用vitamio创建自己的播放器
- 【POJ3268】Silver Cow Party(dij/矩阵转置)
- 排序---堆排序
- HDU(1789)Doing Homework again(贪心,找出损失最少分)
- 抗几何攻击数字水印分类总结
- 自定义Android动画---深入理解matrix
- Smack-Android客户端入门一
- Fragment生命周期
- http://www.cnblogs.com/EricaMIN1987_IT/p/3837436.html
- KindEditor的使用和注意事项
- 数据库的学习笔记
- 排序算法的思考和总结(一)-冒泡和选择排序