Algorithm Advanced Guide-Dynamic Programming-Interval dp

1.
Link to the

problem of merging stones. Solution: For merging an interval of length 2, assuming that the interval is a sand pile as the starting point, we use dp[a][a+1] to represent its value, then $dp[a][a+1]=dp[a][a]+dp[a+1]$[a+1]+value[a]+value[a+1].对于一个区间长度为3的,$dp\left[a\right][a+2]=min(dp[a][a]+dp[a+1][a+2],dp[a][a+1]+dp[a+2][a+2])+value[a]+value[a+1]+value[a+3]$
For the following addition, we can use prefix and for preprocessing

Code:

#include <iostream>
#include <math.h>

using namespace std;
const int N = 310;
int a[310];
int dp[N][N];
int main()
{
    
    
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
    
    
		cin >> a[i];
		dp[i][i] = 0;
	}
	for (int i = 1; i <= n; i++)
		a[i] += a[i - 1];

	for (int l = 2; l <= n; l++)
		for (int b = 1; b <= n - l + 1; b++)
		{
    
    
			int tail = b + l - 1;
			if (b != tail)
				dp[b][tail] = 9999999;
			for (int k = b; k < tail; k++)
				dp[b][tail] = min(dp[b][tail], dp[b][k] + dp[k + 1][tail] + a[tail] - a[b - 1]);
		}
	cout << dp[1][n] << endl;
}

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2.Polygon
topic link

Problem solution: For a ring, we usually like to break it off the loop once

, assuming that we break 1 and n, then we find the maximum value of 1~n to merge, assuming that the short is 1 and 2. , Then we find the maximum value of 2 ~ n+1 for merging. Because we are not sure that we are disconnected from there, we have to consider each case. Then this question is transformed into a similar question to the first question, except The way to merge is different.

Code:

#include <iostream>
#include <queue>
#include <math.h>
#include <string.h>
using namespace std;
const int N = 51;
int dian[2 * N];
char bian[2 * N];
int maxs[3 * N][3 * N], mins[3 * N][3 * N];

int suan(int a, char b, int c)
{
    
    
	if (b == 't')
	{
    
    

		return a + c;
	}
	else
	{
    
    
		return a * c;
	}
}

int main()
{
    
    
	int n;
	cin >> n;

	for (int i = 1; i <= n; i++)
	{
    
    
		cin >> bian[i] >> dian[i];
		bian[n + i] = bian[i];
		dian[n + i] = dian[i];
	}
	int most = 0;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= 2 * n - i + 1; j++)
		{
    
    
			if (i == 1)
			{
    
    
				maxs[j][j] = dian[j];
				mins[j][j] = dian[j];
				maxs[n + j][n + j] = dian[j];
				mins[n + j][n + j] = dian[j];
			}

			else
			{
    
    
				maxs[j][j + i - 1] = -32767;
				mins[j][j + i - 1] = 32767;
				for (int k = j + 1; k - j + 1 <= i; k++)
				{
    
    
					int z = suan(maxs[j][k - 1], bian[k], maxs[k][j + i - 1]);
					int x = suan(mins[j][k - 1], bian[k], mins[k][j + i - 1]);
					int c = suan(mins[j][k - 1], bian[k], maxs[k][j + i - 1]);
					int v = suan(maxs[j][k - 1], bian[k], mins[k][j + i - 1]);
					maxs[j][j + i - 1] = max(maxs[j][j + i - 1], max(z, max(x, max(c, v))));
					mins[j][j + i - 1] = min(mins[j][j + i - 1], min(z, min(x, min(c, v))));

					if (i == n)
					{
    
    
						most = max(most, maxs[j][j + n - 1]);
					}
				}
			}
		}
	cout << most << endl;

	for (int i = 1; i <= n; i++)
	{
    
    
		if (maxs[i][i + n - 1] == most)
			cout << i << " ";
	}
}

---

3. Pyramid
topic Link

Solution:

(borrow Author: do not mind a little chart of the number of ac)
Code:

#include<iostream>
#include<math.h>
#include<string>
#include<string.h>
using namespace std;
typedef long long ll;
const int mod=1e9;
const int N=310;
ll dp[N][N];


int main()
{
    
    
    string a;
    cin>>a;
    int n=a.length();
    for(int i=1;i<=a.length();i+=2)
        for(int l=0;l<=n-i+1;l++)
        {
    
    
           int tail=l+i-1;
           if(i==1)dp[l][l]=1;
           else if(a[l]==a[tail])
           {
    
    
                 for(int k=l;k<tail;k=k+2)
                 {
    
      if(a[k]==a[tail])
                      dp[l][tail]=(dp[l][tail]+dp[l][k]*dp[k+1][tail-1])%mod;

                 }
           }
        }

        cout<<dp[0][a.length()-1]<<endl;
}