Doing Homework again
来源:互联网 发布:知乎钓鱼贴 编辑:程序博客网 时间:2024/06/11 03:59
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#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<iostream>using namespace std;int m,n;struct tv{ int a; int b; double num;} t[1010];int flag[1010];bool cmp(const tv a, const tv b){ if(a.b != b.b) return a.b > b.b; return a.a < b.a;}int main(){ scanf("%d",&m); while(m--) { scanf("%d",&n); memset(flag, 0, sizeof(flag)); int sum1=0,i,j; for(i=0; i<n; i++) scanf("%d",&t[i].a); for(i=0; i<n; i++) { scanf("%d",&t[i].b); } sort(t,t+n,cmp); for( i = 0; i < n; i++) { for(j = t[i].a; j > 0; j--) { if(flag[j] == 0) { flag[j] = 1; break; } } if(j == 0) sum1+= t[i].b; } printf("%d",sum1); printf("\n"); }}排序原理 : 以分数排序 ,若分数相同,则依靠限制天数来排序。先拍分数最大的 尽量排在当天 如果不能排在当天 则向前推 知道零 则表示不能输入了 完成 输出答案
- HDU1789:Doing Homework again
- HDU1789--Doing Homework again
- Doing Homework again
- Doing Homework again(dp)
- hdu1789 Doing Homework again
- HDU Doing Homework again
- hdu Doing Homework again
- Doing Homework again
- 【1789 Doing Homework again】
- HDU1789 Doing Homework again
- Doing Homework again
- HDU1789 Doing Homework again
- Doing Homework again
- Doing Homework again 贪心
- Doing Homework again --贪心
- HDU Doing Homework again
- 【贪心】Doing Homework again
- Doing Homework again
- NYOJ 49-开心的小明:01背包
- -mfloat-abi=softfp的问题,指定fpu为neon
- [转]64位win7系统安装vs2010不成功的解决办法
- 声明与函数、函数指针
- UNIX网络编程——网络IPC:套接字
- Doing Homework again
- Staircases
- iOS利用代理实现界面跳转
- source insight在编辑C时"{"自动缩进的别扭规则的解决.
- UVA 11754 Code Feat 中国剩余定理+普通枚举
- 关系数据库表主键和外键
- SURF源码分析之surflib.h
- Android实现推送方式解决方案
- 博客更新