HDU1074 Doing Homework —— 状压DP
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1074
Doing Homework
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10195 Accepted Submission(s): 4900
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.
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3
Computer
Math
English
3
Computer
English
Math
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.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = +; struct node
{
string name;
int deadline, cost;
}subject[MAXN];
int dp[<<], id[<<], pre[<<];
//id[status]表示当前的状态是由添加哪一科而得来的, id[status]的值即subject数组的下标。
//pre[status]表示当前的状态是由哪一个状态转移过来的。 void Print(int s)
{
if(!s) return;
Print(pre[s]); //追溯上一个状态
cout<< subject[id[s]].name <<endl;
} int main()
{
int T, n;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
for(int i = ; i<n; i++)
cin>>subject[i].name>>subject[i].deadline>>subject[i].cost; memset(dp, -, sizeof(dp));
dp[] = ;
for(int s = ; s<(<<n); s++)
{
int time = ;
for(int i = ; i<n; i++) //计算当前时间
if(s&(<<i))
time += subject[i].cost; for(int i = ; i<n; i++)
{
if(!(s&(<<i)))
{
int exceeded = time + subject[i].cost - subject[i].deadline; //计算超出时间
if(exceeded<) exceeded = ;
int tmp = s|(<<i);
if(dp[tmp]==- || dp[tmp]>dp[s]+exceeded) //更新
{
dp[tmp] = dp[s] + exceeded;
id[tmp] = i;
pre[tmp] = s;
}
}
}
} printf("%d\n", dp[(<<n)-]);
Print((<<n)-);
}
}
HDU1074 Doing Homework —— 状压DP的更多相关文章
- hdu_1074_Doing Homework(状压DP)
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1074 题意:给你n个课程(n<=15)每个课程有限制的期限和完成该课程的时间,如果超出时间,每超 ...
- HDU 1074 Doing Homework 状压dp(第一道入门题)
Doing Homework Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- HDU 1074 Doing Homework (状压DP)
Doing Homework Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- HDU 1074 Doing Homework 状压DP
由于数据量较小,直接二进制模拟全排列过程,进行DP,思路由kuangbin blog得到,膜拜斌神 #include<iostream> #include<cstdio> #i ...
- kuangbin专题十二 HDU1074 Doing Homework (状压dp)
Doing Homework Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- HDU1074:Doing Homework(状压DP)
Doing Homework Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- HDU 1074 Doing Homework【状压DP】
Doing Homework Problem Description Ignatius has just come back school from the 30th ACM/ICPC. Now he ...
- 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 ...
- HDU 1074:Doing Homework(状压DP)
http://acm.hdu.edu.cn/showproblem.php?pid=1074 Doing Homework Problem Description Ignatius has just ...
随机推荐
- pip安装requests库失败
pip install 安装第三方插件是出现Could not fetch URL https://pypi.python.org/simple/pool/: There was a problem ...
- java中String数组和List的互相转化
1.List转String数组 方法一: //先准备一个List List<String> testList=new ArrayList<>(); testList.add(& ...
- XV6锁
锁 xv6 运行在多处理器上,即计算机上有多个单独执行代码的 CPU.这些 CPU 操作同一片地址空间并分享其中的数据结构:xv6 必须建立一种合作机制防止它们互相干扰.即使是在单个处理器上,xv6 ...
- Dream City(线性DP)
描述 JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There are n trees in the ya ...
- 按Esc按钮关闭layer弹窗
//按Esc关闭弹出框 $(document).ready(function () { }).keydown( function (e) { if (e.which === 27) { layer. ...
- POJ-2421Constructing Roads,又是最小生成树,和第八届河南省赛的引水工程惊人的相似,并查集与最小生成树的灵活与能用,水过~~~
Constructing Roads Time Limit: 2000MS Memory Limit: 65536K Description There are N v ...
- POJ 1276 Cash Machine 【DP】
多重背包的模型,但一开始直接将N个物品一个一个拆,拆成01背包竟然T了!!好吧OI过后多久没看过背包问题了,翻出背包九讲看下才发现还有二进制优化一说........就是将n个物品拆成系数:1,2,4, ...
- hihocoder 1873 ACM-ICPC北京赛区2018重现赛 D Frog and Portal
http://hihocoder.com/problemset/problem/1873 时间限制:1000ms 单点时限:1000ms 内存限制:512MB 描述 A small frog want ...
- [Bzoj4817] [Sdoi2017]树点涂色 (LCT神题)
4817: [Sdoi2017]树点涂色 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 629 Solved: 371[Submit][Status ...
- HDU 2059 【DP】
题意: 中文. 思路: 这题不是自己的思想. 当对第i个点的最优值进行求解的时候一定存在最后一个加油的点j.这里j直接枚举. 另外将0和n+1个加油站定义为起点和终点. dp需要加强训练. #incl ...