Niuke Zhou Zhou Lian Qi Rabbit's work (1) (3D dp optimization)

$link$

Ideas:

We set three-dimensional dp: $dp[i][j][k]$到第$I$ have workedjjso far$j$ day, including$I$ worked continuously in$j$ dayskkThe minimum physical strength consumed in$k$ days.

So how to transfer?
When the $When$ you must rest on$i$ day, that is,$s[i]==0$

• $dp[i][j][0]==dp[i-1][j][k]$（其中$1<=k<=j$）;

This iiWhen there is rest or no rest for$i$ days, that is,$s[i]==1$

• 休息：$dp\left[i\right]\left[j\right]\left[0\right]=dp[i-1]\left[j\right]\left[k\right]$（其中$1<=k<=j$）;
• 不休息：$dp[i][j][k]=min(dp[i][j][k],dp[i-1][j-1][k-1]+k)$;

Because of the space limitation of this question, we have to compress the space. Using the compressed space idea of ​​01 backpack, we can remove the first dimension and let $j$ can be traversed in reverse order, and the complexity is$o(n\ast n\ast n)$。

#include<bits/stdc++.h>
using namespace std;

const int maxn = 500,inf = 1e9+20;
int dp[maxn][maxn];
int main()
{
    
    
	int n,k;
	cin >> n >> k;
	string s;
	cin >> s;
	for(int i=0;i<=n+1;i++)
	for(int j=0;j<=n+1;j++)dp[i][j] = inf;
	dp[0][0] = 0;
	for(int i=1;i<=n;i++)
	{
    
    
		for(int j = i; j > 0; j--)
		{
    
    
			for(int k = 1; k <= j; k++)
			{
    
    
				if(s[i-1] == '0')dp[j][0] = min(dp[j][k],dp[j][0]);
				else
				{
    
    
					dp[j][0] = min(dp[j][0],dp[j][k]);
					dp[j][k] = min(dp[j][k] , dp[j-1][k-1] + k);
				}
			}
		}
	}
	for(int i=n;i>0;i--)
	for(int j=1;j<=n;j++)
	if(dp[i][j] <= k)
	{
    
    
		cout<<i<<'\n';
		return 0;
	}
	puts("0");
	return 0;
}