Codeforces Round #323 (Div. 1) B. Once Again... 暴力

B. Once Again...

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/582/problem/B

Description

You are given an array of positive integers a₁, a₂, ..., a_n × T of length n × T. We know that for any i > n it is true that a_i = a_i - n. Find the length of the longest non-decreasing sequence of the given array.

Input

The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 10⁷). The second line contains n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 300).

Output

Print a single number — the length of a sought sequence.

Sample Input

4 33 1 4 2

Sample Output

HINT

题意

给你一个n*t这么长的序列，然后求最长不递减序列

其中a[i+n]=a[i]

题解：

暴力，如果t<=300，我们就直接暴力求就好了

如果t>的话，我们就大胆猜测，中间肯定是连续选一个数

那么我们就预处理前面以a[i]开始的最长，和后面的以a[i]最长是啥就好了~

代码：

#include<iostream>
#include<stdio.h>
#include<queue>
#include<map>
#include<algorithm>
#include<string.h>
using namespace std;#define maxn 3225020int a[maxn];
int dp1[maxn];
int dp2[maxn];
int dp3[maxn];
int lis[maxn];int main()
{int n,t;scanf("%d%d",&n,&t);for(int i=1;i<=n;i++)scanf("%d",&a[i]);if(t<=300){int ans = 0;for(int i=1;i<=n;i++)for(int j=1;j<t;j++)a[j*n+i] = a[i];for(int i=1;i<=n*t;i++){lis[i]=1;for(int j=1;j<i;j++)if(a[i]>=a[j])lis[i]=max(lis[i],lis[j]+1);ans = max(lis[i],ans);}printf("%d\n",ans);return 0;}int k = 200;for(int i=1;i<=n;i++)dp2[a[i]]++;for(int i=1;i<=n;i++)for(int j=1;j<=k;j++)a[j*n+i] = a[i];for(int i=1;i<=k*n;i++){lis[i]=1;for(int j=1;j<i;j++)if(a[i]>=a[j])lis[i]=max(lis[i],lis[j]+1);dp1[a[i]] = max(dp1[a[i]],lis[i]);}memset(lis,0,sizeof(lis));reverse(a+1,a+1+k*n);for(int i=1;i<=k*n;i++){lis[i]=1;for(int j=1;j<i;j++)if(a[i]<=a[j])lis[i]=max(lis[i],lis[j]+1);dp3[a[i]] = max(dp3[a[i]],lis[i]);}int ans = 0;for(int i=0;i<=300;i++)for(int j=i;j<=300;j++)for(int m=j;m<=300;m++)ans = max(dp1[i]+dp2[j]*(t-2*k)+dp3[m],ans);printf("%d\n",ans);
}

转载于:https://www.cnblogs.com/qscqesze/p/4854607.html