#Codeforces 323 [div2] D. Once Again 【优化dp】

最新推荐文章于 2024-07-25 20:12:28 发布

ReskiP

最新推荐文章于 2024-07-25 20:12:28 发布

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分类专栏：解题报告文章标签： dp 算法 codeforces 动态规划优化

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题目

Once Again

PROBLEM

You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - 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, ..., an (1 ≤ ai ≤ 300).

Output

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

        Sample test(s) 
      
          input 
        
4 3
3 1 4 2

          output 
        
5

Note

The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence.

题目意为，100个随机数循环 T 次形成一个数列，求该数列的最长不递减子数列。

循环节T数量级较大，不可能直接DP遍历所有的数字。对于每一节，在DP时只需要对本节自己之前以及上一节所有的数字比较（每个循环节至少会增加一个数字）。若T小于100，那么在T个循环内必然会出现结果，直接DP即可。在T较大时，只有在每个循环节增加数字大于1时才会出现选择的情况。在T等于100时必然会成为稳定状态，之后为将数字最大化，把循环中次数最多的数字插入在剩余的循环节内。

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	long long int dp[105][105];
	static int data2[305];
	int data[105];
	static int maxn,n,t,mmx;
	static long long int ans;
	cin >> n >> t;
	mmx = min(n*n, n*t);

	for (size_t i = 0; i < n; i++)
	{
		cin >> data[i];
		data2[data[i]]++;
	}
	for (size_t i = 0; i < 305; i++)
	{
		maxn = max(maxn, data2[i]);
	}
	for (size_t i = 0; i <mmx/n; i++)
	{
		for (size_t s = 0; s < n; s++)
		{
			for (size_t q = 0; q < n; q++)
			{
				if (i==0)
				{
					dp[i][s] = 1;
					break;
				}
				if (data[s]>=data[q])
				{
					dp[i][s] = max(dp[i][s], dp[i - 1][q]+1);
				}
			}
			for (size_t q = 0; q < s; q++)
			{
				if (data[s] >= data[q])
				{
					dp[i][s] = max(dp[i][s], dp[i][q]+1);
				}
			}
		}
		
	}

	for (size_t i = 0; i < n; i++)
	{
		ans = max(ans, dp[mmx / n - 1][i]);
	}


	cout << ans + maxn*(t - mmx / n);
	return 0;
}