hdu 5064 Find Sequence

 

Find Sequence

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 279    Accepted Submission(s): 80

Problem Description

Give you an positive integer sequence  a1,a2,…,ai,…,an, and they satisfy a1+a2+…+ai+…+an=M(0<M≤222).
We can find new sequence b1=aid1,b2=aid2,…,bx=aidx,…,by=aidy,…,bt=aidt, where if x != y then idx!=idy. and this sequence satisfy:
(1) b1≤b2≤…≤bt 
(2) b2−b1≤b3−b2≤⋯≤bt−bt−1
We can find many sequences b1,b2,b3,…,bt. But we only want to know maximum t.

 

 

Input

The first line in the input file is an Integer  T(1≤T≤30).
The first line of each test case contains two integer n,M(0<M≤222).
Then a line have n integer, they represent a1,a2,…,ai,…,an.

 

 

Output

For each test case, output the maximum t.

 

 

Sample Input

2 6 19 3 2 1 3 4 6 1 4194304 4194304

 

 

Sample Output

5 1

Hint

For the first testcase, The Sequence is 1 2 3 4 6

 

 

Source

BestCoder Round #13

 

题意: 在一个数列里面取出一些数，满足一些条件

官方题解

首先考虑解的结构一定是C1,C1,…,C1,C2,C3,…,Cm这种形式，其中满足C1<C2<C3<…<Cm
所以对a1,a2,a3,…,an去重后从小到大排序得到c1,c2,c3,…,cx其中x是sqrt(M)级别的，用DP[i][j]表示以ci和cj结尾的满足条件的最长序列 首先初值化 DP[i][i]=count(ci) 即ci在原序列中的个数。 而dp[i][j]=max(dp[k][i] 其中k≤i还满足ci−ck≤cj−ci)+1 这样的复杂度是 O(x^3),在题中x最大为1000级别所以会超时，要使用下面优化 因为 dp[i][j]=max(dp[k][i] 其中k≤i还满足ci−ck≤cj−ci)+1 dp[i][j+1]=max(dp[k][i] 其中k≤i还满足ci−ck≤cj+1−ci)+1 注意到cj+1>cj 所以满足ci−ck≤cj−ci的dp[k][i]必然满足ci−ck≤cj+1−ci因而不必重复计算 即最后复杂度可以为O(x^2).

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define maxn (1<<22)+10
using namespace std;

int cnt[maxn],c[2010],dp[2010][2010];
int main()
{
    int i,n,m,j,k;
    int tmp,T,x,ans;
    cin >>T ;
    while(T--)
    {
        scanf("%d%d",&n,&m) ;
        memset(cnt,0,sizeof(cnt));
        for( i = 1 ; i <= n ;i++)
        {
            scanf("%d",&x) ;
            cnt[x]++;
        }
        n = 0 ;
        memset(dp,0,sizeof(dp));
        ans=0;
        for( i =0 ; i <= m ;i++)if(cnt[i])
        {
            c[++n]=i;
            dp[n][n]=cnt[i];
            ans=max(cnt[i],ans);
        }
        for( i = 1 ; i <= n ;i++)
        {
            tmp=dp[i][i];
            k = i ;
            for( j = i+1 ; j <= n ;j++)
            {
                for( ; k >= 1 ;k--)
                {
                    if(c[j]-c[i] >= c[i]-c[k])
                        tmp = max(tmp,dp[k][i]+1) ;
                    else break ;
                }
                dp[i][j]=tmp;
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0 ;
}

View Code

 

转载于:https://www.cnblogs.com/20120125llcai/p/4032336.html