Problem - 1789 继续贪心.经典贪心算法,如果数据比较大就要用线段树来维护了. 思路很简单,只要按照代价由大到小排序,然后靠后插入即可.RE了一次,是没想到deadline可以很大.如果deadline比任务总量要大,显然这个任务是能做的,直接加上去. 代码如下: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <v…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 /*Doing Homework again Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7903 Accepted Submission(s): 4680 Problem Description Ignatius has just c…

Doing Homework again http://acm.hdu.edu.cn/showproblem.php?pid=1789 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 I…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem DescriptionIgnatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Eve…

Doing Homework again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17622    Accepted Submission(s): 10252 Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1789 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…

Doing Homework again 这只是一道简单的贪心,但想不到的话,真的好难,我就想不到,最后还是看的题解 [题目链接]Doing Homework again [题目类型]贪心 &题意: Ignatius有N项作业要完成.每项作业都有限期,如果不在限期内完成作业,期末考就会被扣相应的分数.给出测试数据T表示测试数,每个测试以N开始(N为0时结束),接下来一行有N个数据,分别是作业的限期,再有一行也有N个数据,分别是若不完成次作业会在期末时被扣的分数.求出他最佳的作业顺序后被扣的最小的…

在我上一篇说到的,就是这个,贪心的做法,对比一下就能发现,另一个的扣分会累加而且最后一定是把所有的作业都做了,而这个扣分是一次性的,所以应该是舍弃扣分小的,所以结构体排序后,往前选择一个损失最小的方案直接交换就可以了. #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> using namespace std; struct HomeWork { int de…

意甲冠军  参加大ACM竞争是非常回落乔布斯  每一个工作都有截止日期   未完成必要的期限结束的期限内扣除相应的积分   求点扣除的最低数量 把全部作业按扣分大小从大到小排序  然后就贪阿  能完毕前面的就完毕前面的  实在不能的就扣分吧~ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 1005; int dli[N], red[N],…

Doing Homework again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2969    Accepted Submission(s): 1707 Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he ha…