数列变成一样的时间

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

class solution 
{
public:

	void readdata()
	{
		cin >> n >> m;
		cin.get();

		char ch = ' ';
		while (ch!='\n')
		{
			int e;
			cin >> e;
			S1.push_back(e);
			ch = cin.get();	
		}

		ch = ' ';
		while (ch != '\n')
		{
			int e;
			cin >> e;
			S2.push_back(e);
			ch = cin.get();
		}
	}
	
	void min_time()
	{
		readdata();
		memo = vector<vector<int>>(n, vector<int>(m, -1));

		cout << dp(n - 1, m - 1);
	}
	
private:
	int n;
	int m;
	vector<int> S1;
	vector<int> S2;
	vector<vector<int>> memo;

	int dp(int i, int j)
	{

		if (i==-1)
		{
			int t = 0;
			for (int m = j; m >= 0; m--)
				t += abs(S2[m]);
			return t;
		}

		if (j == -1)
		{
			int t = 0;
			for (int m = i; m >= 0; m--)
				t += abs(S1[m]);
			return t;
		}

		if (memo[i][j]!=-1)
		{
			return memo[i][j];
		}
		int res;

		if (S1[i]==S2[j])
		{
			res = dp(i - 1, j - 1);

		}
		else
		{
			int res1, res2;
			{//操作数列1
				int r1 = dp(i - 1, j - 1) + abs(S1[i] - S2[j]);//替换
				int r2 = dp(i - 1, j) + abs(S1[i]);//删除
				res1 = min(r1, r2);
			}

			{//操作数列2
				int r1 = dp(i - 1, j - 1) + abs(S1[i] - S2[j]);//替换
				int r2 = dp(i, j-1) + abs(S2[j]);//删除
				res2 = min(r1, r2);
			}

			res = min(res1, res2);
			memo[i][j] = res;
		}

		return res;
	}

};

int main()
{
	auto* s = new solution;
	s->min_time();
	return 0;
}