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…

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 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…

原题直通车:HDU  1074  Doing Homework 题意:有n门功课需要完成,每一门功课都有时间期限t.完成需要的时间d,如果完成的时间走出时间限制,就会被减 (d-t)个学分.问:按怎样的顺序做才能使得学分减得最少. 分析:因为n<=15数据比较小,可以用状态DP做.状态k(若k&(1<<j)==1表示第j门功课已经完成,反之未完成), 状态数最多为(1<<16)-1,每个状态k可由状太r(r<k, r&(1<<j)==0且k&…

HDU 1074 Doing Homework (动态规划,位运算) 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 d…

hdu 4825 Xor Sum(trie+贪心) 刚刚补了前天的CF的D题再做这题感觉轻松了许多.简直一个模子啊...跑树上异或x最大值.贪心地让某位的值与x对应位的值不同即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define CLR(a,b) memset((a),(b),sizeof(…

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 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…

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…