hdu 1789 整理下水题 贪心
来源:互联网 发布:贪吃蛇随机算法原理 编辑:程序博客网 时间:2024/04/27 18:14
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
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct Node{ int d,s;}node[1001];int cmp(Node a,Node b){ if(a.s==b.s) { return a.d<b.d; } return a.s>b.s;} int main(){ int used[10000];int T; int n; int j; scanf("%d",&T); while(T--) { scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&node[i].d); for(int i=0;i<n;i++) scanf("%d",&node[i].s); sort(node,node+n,cmp); memset(used,false,sizeof(used)); int ans=0; for(int i=0;i<n;i++) { for(j=node[i].d;j>0;j--) { if(!used[j]) { used[j]=true; break; } } if(j==0) ans+=node[i].s; } printf("%d\n",ans); } return 0;}
0 0
- hdu 1789 整理下水题 贪心
- hdu 1009 整理下水题 贪心
- hdu 2037 整理下水题
- hdu 1870 整理下水题
- hdu 1711 整理下水题
- hdu 1106 整理下水题 排序
- hdu 2020 整理下水题 排序
- hdu 1425 整理下水题 排序
- hdu 1785 整理下水题 排序
- hdu 2602 整理下水题 背包
- hdu 1702 整理下水题 栈和队列
- hdu 1022 整理下水题 栈和队列
- hdu 2013 整理下水题 递推动规
- hdu 2041 整理下水题 递推动规
- hdu 2569 整理下水题 递推动规
- hdu 2084 整理下水题 递推动规
- hdu 1159 整理下水题 递推动规
- HDU-1076-An Easy Task(Debian下水题测试.....)
- <jsp:useBean class="" >
- 分享一个实现图片圆角,圆的自定义的ImageView,尽可能少的减少内存消耗。
- 反转二叉树(二叉树的镜像)
- cocos2dx学习笔记:CCArray-removeObjectAtIndex
- 三层B/S结构
- hdu 1789 整理下水题 贪心
- j2se学习笔记-java.lang中的object类
- LeetCode 116 Populating Next Right Pointers in Each Node
- ArrayList和Linkedlist的区别
- 激光雷达学习笔记(四)定位
- linux程序设计——文件锁定(第七章)
- 最佳替换算法模拟
- RDD特性与操作
- mac下apache和php的简要记录