[dynamic planning -- the best period for buying and selling stocks Series 2]

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Given an array of integers , Among them the first i Elements represent the i Day's share price .
Design an algorithm to calculate the maximum profit . Under the following constraints , You can do as many deals as you can （ Buy and sell a stock many times ）:
You can't participate in multiple transactions at the same time （ You have to sell the stock before you buy it again ）.
After sale of shares , You can't buy shares the next day ( That is, the freezing period is 1 God ).
example

 Input ：[1,2,3,0,2]
 Output ：3
 explain ： The corresponding transaction status is : [ purchase ,  sell ,  Freezing period ,  purchase ,  sell ]

Dynamic programming 5 Step curve

• determine dp The meaning of arrays and subscripts
  dp[i][j], The first i Day status is j, The maximum cash left is dp[i][j].
  The state of this question is relative to A series of It's a lot more complicated . Specifically, it is divided into the following four states ：

 -  State one ： Buy stock status （ Buy stocks today , Or you bought stocks before and didn't operate ）
 
     Status of selling shares , There are two states of selling stocks 
    
 -  State two ： Sold the stock two days ago , Passed the freezing period , Never operated , Keep selling stocks today 
 -  State three ： I sold shares today 

 -  State four ： Today is in a frozen state , But the frozen state is unsustainable , Only one day ！

Notice every state here , For example, state one , The status of buying stocks does not mean that you have bought stocks today , It means to keep the status of buying stocks, that is ： It may have been bought a few days ago , I haven't operated since , So keep buying stocks .

j The status of is ：
0： State one ;1： State two ;2： State three ;3： State four 

Determine the recurrence formula
Reach the status of buying stocks （ State one ） namely ：dp[i][0], There are two specific operations ：

 Operation 1 ： The day before, I was holding stocks （ State one ）,dp[i][0] = dp[i - 1][0]
 Operation two ： I bought it today , There are two situations 
 The day before was frozen （ State four ）,dp[i - 1][3] - prices[i]
 The previous day was to keep selling stocks （ State two ）,dp[i - 1][1] - prices[i]
 So operation two takes the maximum value , namely ：max(dp[i - 1][3], dp[i - 1][1]) - prices[i]
 that dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i]);

To keep selling stocks （ State two ） namely ：dp[i][1], There are two specific operations ：

 Operation 1 ： The day before was state two 
 Operation two ： The day before was frozen （ State four ）
dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);

Reach the state of selling stocks today （ State three ）, namely ：dp[i][2] , There's only one operation ：

 Operation 1 ： Yesterday must have been a stock buying state （ State one ）, Sell today 
 namely ：dp[i][2] = dp[i - 1][0] + prices[i];

Reach the freezing state （ State four ）, namely ：dp[i][3], There's only one operation ：

 Operation 1 ： I sold the stock yesterday （ State three ）
dp[i][3] = dp[i - 1][2];

To sum up, we get a recursive formula ：

dp[i][0] = max(dp[i - 1][0], max(dp[i - 1][3], dp[i - 1][1]) - prices[i]);
dp[i][1] = max(dp[i - 1][1], dp[i - 1][3]);
dp[i][2] = dp[i - 1][0] + prices[i];
dp[i][3] = dp[i - 1][2];

dp How to initialize an array
Here we mainly discuss the second 0 How to initialize .
If you are holding shares （ State one ） that ：dp[0][0] = -prices[0], The cash left over from buying stocks is negative .
Keep selling stocks （ State two ）, The first 0 I haven't sold for days dp[0][1] Initialize to 0 Just go .
I sold shares today （ State three ）, Again dp[0][2] Initialize to 0, Because the least benefit is 0, It will never be negative .
Empathy dp[0][3] Also initially 0.
Determine the traversal order
As can be seen from the recursive formula ,dp[i] Depend on dp[i-1], So it's going from front to back .
Give an example to deduce dp Array
With [1,2,3,0,2] For example ,dp The array is as follows ：

The final result is to take State two , State three , And the maximum value of state four .
The code is as follows :

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
class Solution{
    
	public:
		int maxProfit(vector<int> &prices) {
    
			int n = prices.size();
			if (n == 0) return 0;
			vector<vector<int>> dp(n,vector<int>(4,0));
			dp[0][0] = -prices[0];
			for (int i = 1; i < n; i++) {
    
				dp[i][0] = max({
    dp[i-1][0],dp[i-1][1]-prices[i],dp[i-1][3]-prices[i]});
                // Buy stock status 
				dp[i][1] = max(dp[i-1][1],dp[i-1][3]);
                // Sold shares two days ago , Passed the freezing period , There has been no operation , Keep selling stocks today 
				dp[i][2] = dp[i-1][0] + prices[i];
                // I sold shares today 
				dp[i][3] = dp[i-1][2];
                // Today is the freezing period , But the frozen state is unsustainable , Only one day 
			}
			return max({
    dp[n-1][3],dp[n-1][2],dp[n-1][1]});
		}
}; 
int main() {
    
	Solution Q;
	int n;
	cin>>n;
	vector<int> price(n);
	for (int i = 0; i < n; i++) {
    
		cin>>price[i];
	}
	cout<<Q.maxProfit(price);
}