CF417D--- Cunning Gena(序列+像缩进dp)
A boy named Gena really wants to get to the “Russian Code Cup” finals, or at least get a t-shirt. But the offered problems are too complex, so he made an arrangement with his n friends that they will solve the problems for him.
The participants are offered m problems on the contest. For each friend, Gena knows what problems he can solve. But Gena’s friends won’t agree to help Gena for nothing: the i-th friend asks Gena xi rubles for his help in solving all the problems he can. Also, the friend agreed to write a code for Gena only if Gena’s computer is connected to at least ki monitors, each monitor costs b rubles.
Gena is careful with money, so he wants to spend as little money as possible to solve all the problems. Help Gena, tell him how to spend the smallest possible amount of money. Initially, there’s no monitors connected to Gena’s computer.
Input
The first line contains three integers n, m and b (1 ≤ n ≤ 100; 1 ≤ m ≤ 20; 1 ≤ b ≤ 109) — the number of Gena’s friends, the number of problems and the cost of a single monitor.
The following 2n lines describe the friends. Lines number 2i and (2i + 1) contain the information about the i-th friend. The 2i-th line contains three integers xi, ki and mi (1 ≤ xi ≤ 109; 1 ≤ ki ≤ 109; 1 ≤ mi ≤ m) — the desired amount of money, monitors and the number of problems the friend can solve. The (2i + 1)-th line contains mi distinct positive integers — the numbers of problems that the i-th friend can solve. The problems are numbered from 1 to m.
Output
Print the minimum amount of money Gena needs to spend to solve all the problems. Or print -1, if this cannot be achieved.
Sample test(s)
Input
2 2 1
100 1 1
2
100 2 1
1
Output
202
Input
3 2 5
100 1 1
1
100 1 1
2
200 1 2
1 2
Output
205
Input
1 2 1
1 1 1
1
Output
-1
看到m那么小,就直接想到状压dp了,可是这里有一个monitors的限制,不能暴力枚举这个值
能够先把输入数据按每个人的monitors排序,这样从小到大枚举每个人,边递推边记录了答案即可
/*************************************************************************
> File Name: CF417D.cpp
> Author: ALex
> Mail: zchao1995@gmail.com
> Created Time: 2015年03月16日 星期一 12时33分11秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const double pi = acos(-1.0);
const long long inf = (1LL << 60);
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
LL dp[(1 << 20) + 10];
struct node
{
int sta;
int cost;
int num;
}fri[110];
int cmp (node a, node b)
{
return a.num < b.num;
}
int main ()
{
int n, m, b;
while (~scanf("%d%d%d", &n, &m, &b))
{
for (int i = 0; i <= (1 << m); ++i)
{
dp[i] = inf;
}
dp[0] = 0;
for (int i = 1; i <= n; ++i)
{
int cnt;
fri[i].sta = 0;
int x;
scanf("%d%d%d", &fri[i].cost, &fri[i].num, &cnt);
for (int j = 0; j < cnt; ++j)
{
scanf("%d", &x);
fri[i].sta |= (1 << (x - 1));
}
}
LL ans= inf;
sort (fri + 1, fri + 1 + n, cmp);
for (int i = 1; i <= n; ++i)
{
for (int j = 0; j < (1 << m); ++j)
{
dp[j | fri[i].sta] = min (dp[j] + fri[i].cost, dp[j | fri[i].sta]);
}
ans = min (ans, dp[(1 << m) - 1] + (LL)fri[i].num * b);
}
if (ans >= inf)
{
printf("-1\n");
}
else
{
cout << ans << endl;
}
}
return 0;
}
版权声明:本文博客原创文章。博客,未经同意,不得转载。
CF417D--- Cunning Gena(序列+像缩进dp)的更多相关文章
- codeforces 417D. Cunning Gena 状压dp
题目链接 D. Cunning Gena time limit per test 1 second memory limit per test 256 megabytes input standard ...
- Codeforces 417D Cunning Gena(状态压缩dp)
题目链接:Codeforces 417D Cunning Gena 题目大意:n个小伙伴.m道题目,每一个监视器b花费,给出n个小伙伴的佣金,所须要的监视器数,以及能够完毕的题目序号. 注意,这里仅仅 ...
- Codeforces 417 D. Cunning Gena
按monitor排序,然后状压DP... . D. Cunning Gena time limit per test 1 second memory limit per test 256 megaby ...
- HDU 3681 BFS&像缩进DP&二分法
N*M矩阵.从F出发点.走完全部Y点.每个人格开支1电源点,去G点,电池充满,D无法访问.最小的开始问什么时候满负荷可以去完全部Y.Y和G总共高达15一 第一BFS所有的F.Y.G之间的最短距离. 然 ...
- 括号序列(区间dp)
括号序列(区间dp) 输入一个长度不超过100的,由"(",")","[",")"组成的序列,请添加尽量少的括号,得到一 ...
- Cunning Gena CodeForces - 417D
Cunning Gena CodeForces - 417D 题意 先将小伙伴按需要的监视器数量排序.然后ans[i][j]表示前i个小伙伴完成j集合内题目所需最少钱.那么按顺序枚举小伙伴,用ans[ ...
- 区间和序列上的dp
区间上的dp状态设计最基本的形式: \(F[i]\)表示以i结尾的最优值或方案数. \(F[i][k]\)表示以i结尾附加信息为k的最优值或方案数. 当然可以有多维附加信息. 转移的话往往是枚举上一个 ...
- bzoj4032/luoguP4112 [HEOI2015]最短不公共子串(后缀自动机+序列自动机上dp)
bzoj4032/luoguP4112 [HEOI2015]最短不公共子串(后缀自动机+序列自动机上dp) bzoj Luogu 题解时间 给两个小写字母串 $ A $ , $ B $ ,请你计算: ...
- Easy 2048 Again - ZOJ 3802 像缩进dp
Easy 2048 Again Time Limit: 2 Seconds Memory Limit: 65536 KB Dark_sun knows that on a single-tr ...
随机推荐
- UDP vs. TCP
UDP vs. TCP 原文:UDP vs. TCP,作者是Glenn Fiedler,专注于游戏网络编程相关工作多年. 说在最前面的话 翻译这篇文章的初衷:我在工作中根本接触不到网络游戏编程,但是我 ...
- 端口映射工具 redir/socat/xinetd - 运维技术 - 开源中国社区
端口映射工具 redir/socat/xinetd - 运维技术 - 开源中国社区 端口映射工具 redir/socat/xinetd 10人收藏此文章, 我要收藏 发表于3天前(2013-08 ...
- poj2226(最小点覆盖)
传送门:Muddy Fields 题意:一个由r行c列方格组成的田地,里面有若干个方格充满泥泞,其余方格都是草.要用长度不限,宽度为1的长木板来覆盖这些泥方格,但不能覆盖草地.最少要用多少个长木板. ...
- Android 程序静态分析
简介 静态分析是探索Android程序内幕的一种最常见的方法,它与动态调剂双剑合璧,帮助分析人员解决分析时遇到的各种“疑难”问题. 静态分析是指在不运行的情况下,采用词法分析.语法分析等各种技术手段对 ...
- hdu 4925 贪心 自己从小到大做数据找方法规律
http://acm.hdu.edu.cn/showproblem.php?pid=4925 自己逐个做数据找规律.提供下我的找的: 1 2 1 3 2 2 2 3 3 3 然后发现这种矩阵是最优的: ...
- ural1018(树形dp)
题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=17662 题意:给一棵边有权值的二叉树,节点编号为1-n,1是根节点 ...
- Python – Get Object’s Class Name | Ridge Solutions, Ireland
Python – Get Object’s Class Name | Ridge Solutions, Ireland Python – Get Object’s Class Name Author: ...
- Unity MVC框架 StrangeIoC
StrangeIoC是一个超轻量级和高度可扩展的控制反转(IoC)框架,专门为C#和Unity编写. 项目地址:https://github.com/strangeioc/strangeioc 文档地 ...
- 压缩js参数保存解决方法
angular.js 中 找到 if (typeof fn === 'function') 在 if (fn.length) { console.warn("此函数没有注入注解,将导致编译 ...
- Driver 初始化顺序
Linux系统使用两种方式去加载系统中的模块:动态和静态. 静态加载:将所有模块的程序编译到Linux内核中,由do_initcall函数加载 核心进程(/init/main.c)kernel_ini ...