Vladik and cards
2 seconds
256 megabytes
standard input
standard output
Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:
- the number of occurrences of each number from 1 to 8 in the subsequence doesn't differ by more then 1 from the number of occurrences of any other number. Formally, if there are ck cards with number k on them in the subsequence, than for all pairs of integers
the condition |ci - cj| ≤ 1 must hold.
- if there is at least one card with number x on it in the subsequence, then all cards with number x in this subsequence must form a continuous segment in it (but not necessarily a continuous segment in the original sequence). For example, the subsequence[1, 1, 2, 2] satisfies this condition while the subsequence [1, 2, 2, 1] doesn't. Note that [1, 1, 2, 2] doesn't satisfy the first condition.
Please help Vladik to find the length of the longest subsequence that satisfies both conditions.
The first line contains single integer n (1 ≤ n ≤ 1000) — the number of cards in Vladik's sequence.
The second line contains the sequence of n positive integers not exceeding 8 — the description of Vladik's sequence.
Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions.
3
1 1 1
1
8
8 7 6 5 4 3 2 1
8
24
1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8
17
In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition.
分析:状压dp;
dp[i][j]表示到i为止j里面二进制1表示这个位置的数是否用过的取len+1的个数的最大值;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
#define intxt freopen("in.txt","r",stdin)
const int maxn=1e3+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline void umax(int&p,int q){if(p<q)p=q;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t,pos[][maxn],dp[maxn][<<],cur[],a[maxn],ans;
bool ok(int len)
{
memset(dp,-inf,sizeof(dp));
memset(cur,,sizeof(cur));
dp[][]=;
bool flag=false;
for(int i=;i<=n;i++)
{
for(int j=;j<(<<);j++)
{
if(dp[i][j]==-inf)continue;
for(int k=;k<=;k++)
{
if(j&(<<(k-)))continue;
if(cur[k]+len>pos[k][])continue;
int now_pos=pos[k][cur[k]+len];
umax(dp[now_pos][j|(<<(k-))],dp[i][j]);
if(cur[k]+len+>pos[k][])continue;
now_pos=pos[k][cur[k]+len+];
umax(dp[now_pos][j|(<<(k-))],dp[i][j]+);
}
}
cur[a[i]]++;
}
for(int i=;i<=n+;i++)
{
if(dp[i][(<<)-]>=)
{
flag=true;
ans=max(ans,dp[i][(<<)-]*(len+)+(-dp[i][(<<)-])*len);
}
}
return flag;
}
int main()
{
int i,j;
scanf("%d",&n);
rep(i,,n)a[i]=read();
rep(i,,n)
{
pos[a[i]][++pos[a[i]][]]=i;
}
rep(i,,)if(pos[i][])ans++;
int l=,r=n/;
while(l<=r)
{
int mid=l+r>>;
if(ok(mid))l=mid+;
else r=mid-;
}
printf("%d\n",ans);
//system("Pause");
return ;
}
Vladik and cards的更多相关文章
- Codeforces Round #384 (Div. 2) 734E Vladik and cards
E. Vladik and cards time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #384 (Div. 2) E. Vladik and cards 状压dp
E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way ...
- [codeforces743E]Vladik and cards
E. Vladik and cards time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- CF384 div2 E. Vladik and cards
题意 给你一个的排列,求一个满足条件的最长子序列 每种数字的差小于等于,并且每种数字之内是连续的 解法 首先单纯认为用肯定不行的 所以应该考虑二分答案(所求长度具有二分性) 再用dp判断是否可行,这个 ...
- Vladik and cards CodeForces - 743E (状压)
大意: 给定序列, 求选出一个最长的子序列, 使得任选两个[1,8]的数字, 在子序列中的出现次数差不超过1, 且子序列中相同数字连续. 正解是状压dp, 先二分转为判断[1,8]出现次数>=x ...
- CodeForces743E. Vladik and cards 二分+状压dp
这个题我们可以想象成_---___-----__的一个水柱它具有一遍优一遍行的性质因此可以用来二分最小值len,而每次二分后我们都要验根,we可以把这个水柱想成我们在每个数段里取前一段的那个数后一段有 ...
- 【codeforces 743E】Vladik and cards
[题目链接]:http://codeforces.com/problemset/problem/743/E [题意] 给你n个数字; 这些数字都是1到8范围内的整数; 然后让你从中选出一个最长的子列; ...
- Codeforces Round #384 (Div. 2) //复习状压... 罚时爆炸 BOOM _DONE
不想欠题了..... 多打打CF才知道自己智商不足啊... A. Vladik and flights 给你一个01串 相同之间随便飞 没有费用 不同的飞需要费用为 abs i-j 真是题意杀啊, ...
- 「算法笔记」状压 DP
一.关于状压 dp 为了规避不确定性,我们将需要枚举的东西放入状态.当不确定性太多的时候,我们就需要将它们压进较少的维数内. 常见的状态: 天生二进制(开关.选与不选.是否出现--) 爆搜出状态,给它 ...
随机推荐
- hdu3579 Hello Kiki(数论)
用到中国剩余定理,然后用扩展欧几里得算法求解. 这里有两个注意点,1.硬币数量不能为0或者负数 2.每个group数量有可能大于50,样例中就有 #include<stdio.h> #in ...
- java nio io模型
I/O模型 在开始NIO的学习之前,先对I/O的模型有一个理解,这对NIO的学习是绝对有好处的.我画一张图,简单表示一下数据从外部磁盘向运行中进程的内存区域移动的过程: 这张图片明显忽略了很多细节,只 ...
- Java Web学习笔记--JSP for循环
JSP for循环 <%@ page language="java" contentType="text/html; charset=UTF-8" %&g ...
- CSS3的radial-gradient(径向渐变)
所谓径向渐变,如图下,类似光晕 语法: radial-gradient( [ [渐变大小]? [ at 渐变圆心坐标]? ,]? 颜色[ 开始位置] [,颜色[ 开始位置]]+); ...
- Kattis - Aaah!
Aaah! Photo by Unknown Jon Marius shouted too much at the recent Justin Bieber concert, and now need ...
- String.split()分割字符串
string.split(s[, sep[, maxsplit]]) Return a list of the words of the string s. If the optional secon ...
- maven项目添加websocket
最近由于项目业务需求,需要用到websocket来实现即时信息的推送,学习了一下websocket,网上搜了一下学习教程,很多版本看的我云里雾里,最后选择用tomcat提供的最新版本(tomcat 启 ...
- NOIP2016游记(非题解)
去年的比赛现在来发是不是晚了. -------------------------------- Day1-白天 出发啦, 动车购票处一群丧病的又在玩售票机 动车上看到胡神打苍蝇 苍蝇打苍蝇 在车上颓 ...
- Python快捷键
IDLE默认不能显示行号,使用ALT+G 跳到对应行号,在右下角有显示光标所在行.列. ALT+P 上一个历史输入内容. ALT+N 下一个历史输入内容. IDLE中按F5可以运行代码.
- Linux 分区挂载方案
/boot 1G swap 2G(看内存决定) / 10-15G /home 5G