Doing Homework HDU - 1074
InputThe input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework).
Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.
OutputFor each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.
Sample Input
2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3
Sample Output
2
Computer
Math
English
3
Computer
English
Math
Hint
In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the
word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.
第一个DP状态压缩的题,看了好久。希望下次做的时候可以懂一点点吧。
// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#define INF 0x3f3f3f3f
#define mod 2016
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = (<<)+;
int n, m, T, len, cnt, num, ans, Max;
int dp[maxn], t[maxn], pre[maxn], d[], f[];
char str[][]; void output(int x) {
if( !x ) return ;
output(x-(<<pre[x]));
printf("%s\n", str[pre[x]]);
} void input() {
scanf("%d", &T);
while( T -- ) {
scanf("%d", &n);
for(int i=; i<n; i++) scanf("%s%d%d",str[i], &d[i], &f[i]);
int bit = << n;
for(int i=; i<=bit; i++) {
dp[i] = INF;
for(int j=n-; j>=; j--) {
int te = <<j;
if( !(i&te) ) continue;
int sc = t[i-te]+f[j]-d[j];
if( sc< ) sc = ;
if( dp[i]>dp[i-te]+sc) {
dp[i] = dp[i-te]+sc;
t[i] = t[i-te]+f[j];
pre[i] = j;
}
}
}
printf("%d\n", dp[bit-]);
output(bit-);
}
} int main() {
input();
return ;
}
感谢大神http://blog.csdn.net/xingyeyongheng/article/details/21742341
Doing Homework HDU - 1074的更多相关文章
- Doing Homework HDU - 1074 (状压dp)
Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every ...
- D - Doing Homework HDU - 1074 (状压dp)
题目链接:https://cn.vjudge.net/contest/68966#problem/D 具体思路:我们可以把每个情况都枚举出来,然后用递归的形式求出最终的情况. 比如说 我们要求 10 ...
- Doing Homework HDU - 1074 状态压缩
#include<iostream> #include<cstring> #include<cstdio> #include<string> #incl ...
- Day9 - G - Doing Homework HDU - 1074
有n个任务,每个任务有一个截止时间,超过截止时间一天,要扣一个分.求如何安排任务,使得扣的分数最少.Input有多组测试数据.第一行一个整数表示测试数据的组数第一行一个整数n(1<=n<= ...
- 【状态DP】 HDU 1074 Doing Homework
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 ...
- HDU 1074 Doing Homework (动态规划,位运算)
HDU 1074 Doing Homework (动态规划,位运算) Description Ignatius has just come back school from the 30th ACM/ ...
- HDU 1074 Doing Homework (dp+状态压缩)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1074 题目大意:学生要完成各科作业, 给出各科老师给出交作业的期限和学生完成该科所需时间, 如果逾期一 ...
- HDU 1074:Doing Homework(状压DP)
http://acm.hdu.edu.cn/showproblem.php?pid=1074 Doing Homework Problem Description Ignatius has just ...
- HDU 1074 Doing Homework(状压DP)
第一次写博客ORZ…… http://acm.split.hdu.edu.cn/showproblem.php?pid=1074 http://acm.hdu.edu.cn/showproblem.p ...
随机推荐
- contos mysql 删除
yum remove mysql mysql-server mysql-libs compat-mysql51rm -rf /var/lib/mysqlrm /etc/my.cnf查看是否还有mysq ...
- PHP的类,abstract类,interface及关键字extends和implements
原文:https://blog.csdn.net/qq_19557947/article/details/77880757?locationNum=4&fps=1 PHP类 PHP类是单继承, ...
- 通过wui登陆 sap 页面对数据进行高级 搜索
1: 登陆QGL系统. 在 T-CODE搜索框输入wui 会跳到搜索的web页面,进行搜索. 或者浏览器输入: https://ldciqgl.wdf.sap.corp:44300/sap(bD1lb ...
- 微信小程序--修改data数组或对象里面的值
1.初始data数据 Page({ data:{ code:'1234', reward:[{ name:"艾伦" ...
- SpringBoot java.lang.IllegalArgumentException: Request header is too large
在application.properties##tomcat 请求设置server.max-http-header-size=1048576server.tomcat.max-connections ...
- 【Java】-NO.16.EBook.4.Java.1.012-【疯狂Java讲义第3版 李刚】- JDBC
1.0.0 Summary Tittle:[Java]-NO.16.EBook.4.Java.1.012-[疯狂Java讲义第3版 李刚]- JDBC Style:EBook Series:Java ...
- NetBeans issues and solutions.(build.xml and debug multiple projects)
Copy a directory to another directory when building the .jar in NetBeans in the build.xml file. Solu ...
- 包含mysql 递归查询父节点 和子节点
包含mysql 递归查询父节点 和子节点 mysql递归查询,查父集合,查子集合 查子集合 --drop FUNCTION `getChildList` CREATE FUNCTION `getChi ...
- Redis:Sentinel哨兵
简介 Sentinel的作用就是主从切换:Redis-Sentinel是Redis官方推荐的高可用性(HA)解决方案,当用Redis做Master-slave的高可用方案时,假如master宕机了,R ...
- JQuery 获取页面某一元素在屏幕上的位置
获取页面某一元素的绝对X,Y坐标 var X = $('#ElementID').offset().top;//元素在当前视窗距离顶部的位置 var Y = $('#ElementID').offse ...