CoderForces999F-Cards and Joy
2 seconds
256 megabytes
standard input
standard output
There are nn players sitting at the card table. Each player has a favorite number. The favorite number of the jj-th player is fjfj.
There are k⋅nk⋅n cards on the table. Each card contains a single integer: the ii-th card contains number cici. Also, you are given a sequence h1,h2,…,hkh1,h2,…,hk. Its meaning will be explained below.
The players have to distribute all the cards in such a way that each of them will hold exactly kk cards. After all the cards are distributed, each player counts the number of cards he has that contains his favorite number. The joy level of a player equals htht if the player holds tt cards containing his favorite number. If a player gets no cards with his favorite number (i.e., t=0t=0), his joy level is 00.
Print the maximum possible total joy levels of the players after the cards are distributed. Note that the sequence h1,…,hkh1,…,hk is the same for all the players.
The first line of input contains two integers nn and kk (1≤n≤500,1≤k≤101≤n≤500,1≤k≤10) — the number of players and the number of cards each player will get.
The second line contains k⋅nk⋅n integers c1,c2,…,ck⋅nc1,c2,…,ck⋅n (1≤ci≤1051≤ci≤105) — the numbers written on the cards.
The third line contains nn integers f1,f2,…,fnf1,f2,…,fn (1≤fj≤1051≤fj≤105) — the favorite numbers of the players.
The fourth line contains kk integers h1,h2,…,hkh1,h2,…,hk (1≤ht≤1051≤ht≤105), where htht is the joy level of a player if he gets exactly tt cards with his favorite number written on them. It is guaranteed that the condition ht−1<htht−1<ht holds for each t∈[2..k]t∈[2..k].
Print one integer — the maximum possible total joy levels of the players among all possible card distributions.
4 3
1 3 2 8 5 5 8 2 2 8 5 2
1 2 2 5
2 6 7
21
3 3
9 9 9 9 9 9 9 9 9
1 2 3
1 2 3
0
In the first example, one possible optimal card distribution is the following:
- Player 11 gets cards with numbers [1,3,8][1,3,8];
- Player 22 gets cards with numbers [2,2,8][2,2,8];
- Player 33 gets cards with numbers [2,2,8][2,2,8];
- Player 44 gets cards with numbers [5,5,5][5,5,5].
Thus, the answer is 2+6+6+7=212+6+6+7=21.
In the second example, no player can get a card with his favorite number. Thus, the answer is 00.
题意:n∗kn∗k张卡片分给n个人,每人k张。第二行输入n∗kn∗k张卡片上面写的数字,第三行输入nn个人喜欢的数字,第四行输入k个数字,h[i]h[i]表示拿到ii张自己喜欢的卡片可以获得的快乐值。问所有人快乐值之和最大为多少?
题解:DP,dp[i][j],表示i个相同的数字给j个人(这j个人都喜欢这相同的数字);cnt[i]表示数字i的数量,num[j]表示喜欢j 的人数;
状态转换方程为:dp[i][j]=max(dp[i][j],dp[i-u][j-1]+w[u]);(1=<u<=k)
AC代码为:
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n,k,a[maxn],f[5005],w[15];
int cnt[maxn],num[maxn],dp[5005][521];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
memset(cnt,0,sizeof cnt);
memset(num,0,sizeof num);
cin>>n>>k;
for(int i=1;i<=n*k;i++) cin>>a[i],cnt[a[i]]++;
for(int i=1;i<=n;i++) cin>>f[i],num[f[i]]++;
for(int i=1;i<=k;i++) cin>>w[i];
for(int i=1;i<=n*k;i++)
{
dp[i][1]=w[min(i,k)];
for(int j=2;j<=n;j++)
{
for(int u=1;u<=min(i,k);u++)
dp[i][j]=max(dp[i][j],dp[i-u][j-1]+w[u]);
}
}
long long ans=0;
for(int i=1;i<maxn;i++) if(num[i]) ans+=dp[cnt[i]][num[i]];
cout<<ans<<endl;
return 0;
}
CoderForces999F-Cards and Joy的更多相关文章
- Codeforces Round #490 (Div. 3) F - Cards and Joy
F - Cards and Joy 思路:比较容易想到dp,直接dp感觉有点难,我们发现对于每一种数字要处理的情况都相同就是有 i 张牌 要给 j 个人分, 那么我们定义dp[ i ][ j ]表示 ...
- F. Cards and Joy
F. Cards and Joy 题目大意: 给你n个人,每一个人恰好选k张牌. 第一行是 n 和 k 第二行有n*k个数,代表有n*k张牌,每张牌上的数字 第三行有n个数,代表第i个人喜欢的数字 第 ...
- Cards and Joy CodeForces - 999F (贪心+set)
There are nn players sitting at the card table. Each player has a favorite number. The favorite numb ...
- Codeforces 999F Cards and Joy(二维DP)
题目链接:http://codeforces.com/problemset/problem/999/F 题目大意:有n个人,n*k张卡牌,每个人会发到k张卡牌,每个人都有一种喜欢的卡牌f[i],当一个 ...
- Codeforces Round #490 (Div. 3) :F. Cards and Joy(组合背包)
题目连接:http://codeforces.com/contest/999/problem/F 解题心得: 题意说的很复杂,就是n个人玩游戏,每个人可以得到k张卡片,每个卡片上有一个数字,每个人有一 ...
- 999F Cards and Joy
传送门 题目大意 有n个人n*m张牌,每个人分m张牌.每个人有一个自己喜欢的数值,如果他的牌中有x张数值等于这个值则他的高兴度为L[x],求怎样分配牌可以使得所有人的总高兴度最大. 分析 我们发现每一 ...
- Codeforces Round #490 (Div. 3)
感觉现在\(div3\)的题目也不错啊? 或许是我变辣鸡了吧....... 代码戳这里 A. Mishka and Contes 从两边去掉所有\(≤k\)的数,统计剩余个数即可 B. Reversi ...
- [Codeforces]Codeforces Round #490 (Div. 3)
Mishka and Contest #pragma comment(linker, "/STACK:102400000,102400000") #ifndef ONLINE_JU ...
- BZOJ 1004 【HNOI2008】 Cards
题目链接:Cards 听说这道题是染色问题的入门题,于是就去学了一下\(Bunside\)引理和\(P\acute{o}lya\)定理(其实还是没有懂),回来写这道题. 由于题目中保证"任意 ...
随机推荐
- 使用Samba服务实现文件共享
1.在虚拟机上安装Samba服务安装包 (在下载之前检查客户机与服务器是否能够ping通) (鼠标右击桌面,打开终端,测试和yum是否能够ping通,下面的命令行是我的yum的IP地址) [root@ ...
- C#: 统计method的执行时间
对于性能分析来说,无非是内存占用,CPU使用和执行时间. 那么,对于执行时间(elapsed times)的测量,需要强调的是,尽量不要使用DateTime类来,而是应该使用Stopwatch 类.M ...
- Swoole跟thinkphp5结合开发WebSocket在线聊天通讯系统
ThinkPHP使用Swoole需要安装 think-swoole Composer包,前提系统已经安装好了Swoole PECL 拓展* tp5的项目根目录下执行composer命令安装think- ...
- [TCP] TCP协议族的学习 and TCP协议
1.TCP协议族这个大家庭,每个协议在OSI5层模型中所处的位子 其中,网络层里的 ICMP = Internet Control Message Protocol,即因特网控制报文协议, IGMP ...
- thinkphp 获取前端传递过来的参数
thinkphp 获取前端传递过来的参数 use think\facade\Request; // 获取当前请求的name变量 Request::param('name'); // 获取当前请求的所有 ...
- nyoj 7 街区最短路径问题 (曼哈顿距离(出租车几何) or 暴力)
街区最短路径问题 时间限制:3000 ms | 内存限制:65535 KB 难度:4 描述 一个街区有很多住户,街区的街道只能为东西.南北两种方向. 住户只可以沿着街道行走. 各个街道之间的间 ...
- 利用Python学习线性代数 -- 1.1 线性方程组
利用Python学习线性代数 -- 1.1 线性方程组 本节实现的主要功能函数,在源码文件linear_system中,后续章节将作为基本功能调用. 线性方程 线性方程组由一个或多个线性方程组成,如 ...
- SpringBoot Application深入学习
本节主要介绍SpringBoot Application类相关源码的深入学习. 主要包括: SpringBoot应用自定义启动配置 SpringBoot应用生命周期,以及在生命周期各个阶段自定义配置. ...
- Install aws cli
下载 https://s3.amazonaws.com/aws-cli/AWSCLI64PY3.msi 添加环境变量
- 线程中synchronized关键字和lock接口的异同
一.synchronized关键字 1.可以用来修饰代码块 synchronized (this) { // 同步的关键字 this 表示当前线程对象 if (num == 0) { break; } ...