Codeforces Round #384 (Div. 2) E. Vladik and cards 状压dp

题目链接

http://codeforces.com/contest/743/problem/E

题面

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.

输入

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.

输出

Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions.

样例输入

3
1 1 1

样例输出

1

题意

给你n个数字，你需要找到一个最长的子序列，满足以下要求:

1.对于每个i和j，要求abs(num[i]-num[j])<=1，num[i]表示这个数字i出现的次数

2.所有相同的数字应该挨在一起。

求最长的子序列长度

题解

枚举每个数字的长度num，那么显然每个数字要么是num，要么就是num+1

然后我们对于每个长度进行check就好了

dp[i][j]表示当前状态为i的时候，其中有i个数的长度为num+1，用一个next进行转移就好了

next[i][j][k]表示从i开始，j出现k次的位置是啥位置。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>

using namespace std;

int n;
int a[1005];
int nex[1005][10][150];
int dp[1<<10][10];

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=0;i<=1000;i++)
        {
            for(int j=0;j<10;j++)
            {
                for(int k=0;k<150;k++)
                {
                    nex[i][j][k]=1e9;
                }
            }
        }
        for(int i=0;i<(1<<10);i++)
        {
            for(int j=0;j<10;j++)
            {
                dp[i][j]=1e9;
            }
        }
        for(int i=0;i<=n;i++)
        {
            for(int j=1;j<=8;j++)
            {
                int cnt=0;
                nex[i][j][0]=i;
                for(int k=i+1;k<=n;k++)
                {
                    if(a[k]==j)
                    {
                        cnt++;
                        nex[i][j][cnt]=k;
                    }
                }
                /*for(int k=0;k<=10;k++)
                {
                    cout<<i<<":"<<j<<":"<<k<<":"<<nex[i][j][k]<<endl;
                }*/
            }
        }
        int len=(1<<8);
        int ans=0;
        dp[0][0]=0;
        for(int num=0;num*8<=n;num++)
        {
            for(int i=0;i<len;i++)
                for(int j=0;j<=8;j++)
                    dp[i][j]=1e9;
            dp[0][0]=0;
            for(int i=0;i<len;i++)
            {
                for(int j=0;j<=8;j++)
                {
                    if(dp[i][j]==1e9)
                        continue;

                    for(int k=1;k<=8;k++)
                    {
                        int id=(1<<(k-1));
                        if((i&id)==0)
                        {
                            dp[i|id][j]=min(dp[i|id][j],nex[dp[i][j]][k][num]);
                            dp[i|id][j+1]=min(dp[i|id][j+1],nex[dp[i][j]][k][num+1]);
                        }
                    }
                }
            }
            for(int i=0;i<=8;i++)
            {
                if(dp[len-1][i]<=n)
                    ans=max(ans,8*num+i);
            }
            //cout<<num<<" : "<<ans<<endl;
        }
        printf("%d\n",ans);
    }
}