Dynamic Programming and Applications

Dynamic Programming

And the largest contiguous subsequence

The basic method for solving

A method
DP [i] denotes the A [1] and the maximum contiguous subsequence A [i] of the
final maximum and contiguous subsequence = dp [num]

#include<iostream>

using namespace std;

const int N = 10000;
int A[N];
int dp[N];

int main()
{
	int num = 6;
	for (int i = 1; i <= num; i++)
		cin >> A[i];
	int temp = -999999;
	int sum = 0;
	for (int i = num; i >= 1; i--)
	{
		sum += A[i];
		if (sum > temp)
			temp = sum;
		dp[i] = temp;
		if (sum < 0)
			sum = 0;
	}
	printf("%d", dp[1]);
	system("pause");
	return 0;

}

Method Two
dp [I] represents a maximum and A [i] at the end of the contiguous sequence
determined dp entire array, through the entire array and then find the maximum value is the greatest contiguous subsequences and

3 movable normalized points:
state variable: dp [i]
Boundary: dp [1] = A [ 1]
a state transition equation: dp [i] = max ( A [i], dp [i-1] + A [i ])

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 10000;
int A[N];
int dp[N];

int main()
{
	int num = 6;
	for (int i = 1; i <= num; i++)
		cin >> A[i];
	//边界
	dp[1] = A[1];

	for (int i = 2; i <=num; i++)
	{
		dp[i] = max(dp[i - 1] + A[i], A[i]);
	} 
	int k = dp[1];
	for (int i = 2; i <=num  ; i++)
	{
		if (dp[i] > k)
			k = dp[i];
	}
	printf("%d",k);
	system("pause");
	return 0;
}

Example 1 Maximum sum

Source : 2479 the Maximum SUM

#include<iostream>

using namespace std;
const int N = 100010;
int A[N];
int dpr[N];		//从右向左dp,求出连续子序列的最大和。dp[i]表示从A[1]到A[i]中连续子序列的最大和
int dpl[N];		//从左向右dp,求出连续子序列的最大和。dp[i]表示从A[i]到A[N]中连续子序列的最大和

int main()
{
	int T,k;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &k);
		//将整个序列读入
		for (int i = 1; i <= k; i++)
			cin >> A[i];
		
		//求解dpr[i]
		int sum = 0;
		int temp = -9999999;
		for (int i = 1; i <= k; i++)
		{
			sum += A[i];
			if (sum > temp)
				temp = sum;
			dpr[i] = temp;
			if (sum < 0)
				sum = 0;
		}
		sum = 0; temp = -9999999;

		//求解dpl[i]
		for (int i = k; i >=1; i--)
		{
			sum += A[i];
			if (sum > temp)
				temp = sum;
			dpl[i] = temp;
			if (sum < 0)
				sum = 0;
		}
		int MaxSum = -9999999;
		for (int i = 1; i < k; i++)
		{
			int t = dpr[i] + dpl[i + 1];
			if (MaxSum < t)
				MaxSum = t;
		}
		cout << MaxSum << endl;
	}
	system("pause");
	return 0;
}

Best plus expression

example

POJ 4103 stepped squares

Ideas: ways (i, j, n): A method from node (i, j) began to walk step number n

ways(i,j,n) = ways(i,j-1,n-1) + ways(i,j+1,n-1) + ways(i+1,j,n-1)

Use array vis [i] [j] labeled node (i, j) whether through

Tips：

1. N is 0 or not is determined first, then the node (i, j) is labeled

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int vis[50][50];

int ways(int i, int j, int n)
{
	if (n == 0)
	{
		return 1;
	}
	int num = 0;
	if (!vis[i][j - 1])
	{
		vis[i][j - 1] = 1;
		num += ways(i, j - 1, n - 1);
		vis[i][j - 1] = 0;
	}
	if (!vis[i][j + 1])
	{
		vis[i][j + 1] = 1;
		num += ways(i, j + 1, n - 1);
		vis[i][j + 1] = 0;
	}
	if (!vis[i+1][j])
	{
		vis[i + 1][j] = 1;
		num += ways(i+1, j, n - 1);
		vis[i + 1][j] = 0;
	}
	return num;
}

int main()
{
	int n;
	cin >> n;
	memset(vis, 0, sizeof(vis));
	vis[0][25] = 1;
	cout << ways(0, 25, n) << endl;
	return 0;
}