给定时间后完全装满的容器数

给定两个分别表示级别数和秒数的整数N和T ，任务是在给定条件下找到T秒后完全填充的容器数：

• N级血管的结构使得每一级的血管数等于级数，即1, 2, 3, … 直到 N 。
• 每个容器最多可储存 1 个单位的水，并且每秒钟就有 1 个单位的水以恒定速率从水龙头中倒出。
• 如果容器变满，则水开始从容器中流出，并从容器边缘倾泻而下，并均匀分布在紧邻其下方的两个相连容器中。

假设：

1. 所有对象沿水平轴对称排列。
2. 所有级别均等间隔。
3. 水对称地流过容器的两个边缘。

例子：

Input: N = 3, T = 2
Output: 1
Explanation:
View of Structure with N = 3 and at a time T = 2 after the tap has been opened

Input: N = 3, T = 4
Output: 3
Explanation:
View of Structure with N = 3 and at a time T = 4 after the tap has been opened

朴素的方法：解决问题的最简单方法是检查是否可以在T秒内完全填充x 个容器。如果发现为真，则检查x+1血管并重复以获取x的最大值。
时间复杂度： O(N ³ )
辅助空间： O(1)

高效方法：上述方法可以使用动态规划进行优化。请按照以下步骤解决问题：

1. 将血管结构存储在矩阵中，例如M ，其中M[i][j]表示^第i 层中的^第j^个血管。
2. 对于任何船只M[i][j] ，在紧接较低级别的连接船只是M[i + 1][j]和M[i + 1][j + 1] 。
3. 最初，将所有水放入第一个容器中，即。 M[0][0] = t 。
4. 从最上面的容器 i 开始，每单位时间增量重新计算矩阵的状态，即。 M[0][0] = t 。
5. 如果水量超过容器的体积，则从容器中流下的水量分成 2 等分
6. 部分填充两个相连的容器在立即较低的水平。

C++

// C++ program to implement 
// the above approach 
#include  
using namespace std; 
   
int n, t;
   
// Function to find the number of
// completely filled vessels
int FindNoOfFullVessels(int n, int t)
{
       
    // Store the vessels
    double Matrix[n][n];
   
    // Assuming all water is present
    // in the vessel at the first level
    Matrix[0][0] = t * 1.0;
   
    // Store the number of vessel
    // that are completely full
    int ans = 0;
   
    // Traverse all the levels
    for(int i = 0; i < n; i++)
    {
           
        // Number of vessel at each
        // level is j
        for(int j = 0; j <= i; j++)
        {
               
            // Calculate the exceeded
            // amount of water
            double exceededwater = Matrix[i][j] - 1.0;
   
            // If current vessel has
            // less than 1 unit of
            // water then continue
            if (exceededwater < 0)
                continue;
   
            // One more vessel is full
            ans++;
   
            // If left bottom vessel present
            if (i + 1 < n)
                Matrix[i + 1][j] += exceededwater / 2;
   
            // If right bottom vessel present
            if (i + 1 < n && j + 1 < n)
                Matrix[i + 1][j + 1] += exceededwater / 2;
        }
    }
    return ans;
}
   
// Driver Code
int main()
{
       
    // Number of levels
    int N = 3;
   
    // Number of seconds
    int T = 4;
   
    // Function call
    cout << FindNoOfFullVessels(N, T) << endl;
       
    return 0;
}
   
// This code is contributed by sanjoy_62

Java

// Java Program to implement
// the above approach
   
import java.io.*;
import java.util.*;
   
class GFG {
   
    static int n, t;
   
    // Function to find the number of
    // completely filled vessels
    public static int
    FindNoOfFullVessels(int n, int t)
    {
        // Store the vessels
        double Matrix[][]
            = new double[n][n];
   
        // Assuming all water is present
        // in the vessel at the first level
        Matrix[0][0] = t * 1.0;
   
        // Store the number of vessel
        // that are completely full
        int ans = 0;
   
        // Traverse all the levels
        for (int i = 0; i < n; i++) {
   
            // Number of vessel at each
            // level is j
            for (int j = 0; j <= i; j++) {
   
                // Calculate the exceeded
                // amount of water
                double exceededwater
                    = Matrix[i][j] - 1.0;
   
                // If current vessel has
                // less than 1 unit of
                // water then continue
                if (exceededwater < 0)
                    continue;
   
                // One more vessel is full
                ans++;
   
                // If left bottom vessel present
                if (i + 1 < n)
                    Matrix[i + 1][j]
                        += exceededwater / 2;
   
                // If right bottom vessel present
                if (i + 1 < n && j + 1 < n)
                    Matrix[i + 1][j + 1]
                        += exceededwater / 2;
            }
        }
   
        return ans;
    }
   
    // Driver Code
    public static void main(String[] args)
    {
        // Number of levels
        int N = 3;
   
        // Number of seconds
        int T = 4;
   
        // Function call
        System.out.println(
            FindNoOfFullVessels(N, T));
    }
}

Python3

# Python3 program to implement    
# the above approach    
      
# Function to find the number of    
# completely filled vessels    
def FindNoOfFullVessels(n, t) :    
         
    # Store the vessels    
    Matrix = [[0 for i in range(n)] for j in range(n)]   
      
    # Assuming all water is present    
    # in the vessel at the first level    
    Matrix[0][0] = t * 1.0    
      
    # Store the number of vessel    
    # that are completely full    
    ans = 0    
      
    # Traverse all the levels    
    for i in range(n) :   
             
        # Number of vessel at each    
        # level is j    
        for j in range(i + 1) :   
                 
            # Calculate the exceeded    
            # amount of water    
            exceededwater = Matrix[i][j] - 1.0    
      
            # If current vessel has    
            # less than 1 unit of    
            # water then continue    
            if (exceededwater < 0) :   
                continue   
      
            # One more vessel is full    
            ans += 1    
      
            # If left bottom vessel present    
            if (i + 1 < n) :    
                Matrix[i + 1][j] += exceededwater / 2   
      
            # If right bottom vessel present    
            if (i + 1 < n and j + 1 < n) :   
                Matrix[i + 1][j + 1] += exceededwater / 2    
    return ans   
      
         
# Number of levels    
N = 3   
      
# Number of seconds    
T = 4   
      
# Function call    
print(FindNoOfFullVessels(N, T))
 
# This code is contributed by divyesh072019

C#

// C# program to implement 
// the above approach 
using System; 
   
class GFG{
   
//static int n, t;
   
// Function to find the number of
// completely filled vessels
public static int FindNoOfFullVessels(int n, 
                                      int t)
{
       
    // Store the vessels
    double[,] Matrix = new double[n, n];
   
    // Assuming all water is present
    // in the vessel at the first level
    Matrix[0, 0] = t * 1.0;
   
    // Store the number of vessel
    // that are completely full
    int ans = 0;
   
    // Traverse all the levels
    for(int i = 0; i < n; i++)
    {
   
        // Number of vessel at each
        // level is j
        for(int j = 0; j <= i; j++) 
        {
   
            // Calculate the exceeded
            // amount of water
            double exceededwater = Matrix[i, j] - 1.0;
   
            // If current vessel has
            // less than 1 unit of
            // water then continue
            if (exceededwater < 0)
                continue;
   
            // One more vessel is full
            ans++;
   
            // If left bottom vessel present
            if (i + 1 < n)
                Matrix[i + 1, j] += exceededwater / 2;
   
            // If right bottom vessel present
            if (i + 1 < n && j + 1 < n)
                Matrix[i + 1, j + 1] += exceededwater / 2;
        }
    }
    return ans;
}
   
// Driver Code
public static void Main()
{
       
    // Number of levels
    int N = 3;
   
    // Number of seconds
    int T = 4;
   
    // Function call
    Console.WriteLine(FindNoOfFullVessels(N, T));
}
}
   
// This code is contributed by sanjoy_62

输出：

3

时间复杂度： O(N ² )
空间复杂度： O(N ² )

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