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<M222).
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) b1b2bt 
(2) b2b1b3b2btbt1
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(1T30).
The first line of each test case contains two integer n,M(0<M222).
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
 
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cicj结尾的满足条件的最长序列
首先初值化 DP[i][i]=count(ci)ci在原序列中的个数。
而dp[i][j]=max(dp[k][i] 其中ki还满足cickcjci)+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ki还满足cickcjci)+1
dp[i][j+1]=max(dp[k][i] 其中ki还满足cickcj+1ci)+1
注意到cj+1>cj 所以满足cickcjci的dp[k][i]必然满足cickcj+1ci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

 



posted @ 2014-10-18 00:20  _log__  阅读(138)  评论(0编辑  收藏  举报