算法笔记--动态规划

求解问题类型：重叠子问题

两种写法:

递推：

#include <iostream>
#include <vector>

using namespace std;

const int maxn = 100;
int dp[maxn] = { 0 };

int f_seque(int n)
{
	if (n == 1 || n == 2)
		return 1;
	if (n > 2)
	{
		if (dp[n] == 0)
		{
			dp[n] = f_seque(n - 1) + f_seque(n - 2);
			return dp[n];
		}
		else return dp[n];
	}
}

int main()
{
	int n = f_seque(3);
	cout << n;
	system("pause");
	return 0;
}

递推：

 

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

const int maxn = 100;
int datas[maxn][maxn] = { 0 };
int dp[maxn][maxn] = { 0 };
int main()
{
    int n;
    cin >> n;
    for(int i=1;i<=n;++i)
        for (int j = 1; j <=i; ++j)
        {
            int tem;
            cin >> tem;
            datas[i][j] = tem;
        }
    for (int i = 1; i <= n; ++i)
        dp[n][i] = datas[n][i];
    for(int i=n-1;i>=1;--i)
        for (int j = 1; j <= i; ++j)
        {
            dp[i][j] = max(dp[i + 1][j], dp[i + 1][j + 1]) + datas[i][j];
        }
    cout << dp[1][1];
    system("pause");
    return 0;
}