公路广告牌问题

考虑一条M英里的高速公路。任务是在高速公路上放置广告牌，以使收入最大化。广告牌的可能位置由数字x ₁ < x ₂ < ….. < x _n-1 < x _{n 给出}，指定从道路一端测量的英里位置。如果我们在位置 x _i放置一个广告牌，我们会收到r _i > 0的收入。有一个限制，即在t英里或小于它的范围内不能放置两个广告牌。
注意：所有可能的位置从 x ₁到 x _n都在 0 到 M 的范围内，因为需要在 M 英里的高速公路上放置广告牌。
例子：

Input : M = 20
        x[]       = {6, 7, 12, 13, 14}
        revenue[] = {5, 6, 5,  3,  1}
        t = 5
Output: 10
By placing two billboards at 6 miles and 12
miles will produce the maximum revenue of 10.

Input : M = 15
        x[] = {6, 9, 12, 14}
        revenue[] = {5, 6, 3, 7}
        t = 2
Output : 18  

设 maxRev[i], 1 <= i <= M，是高速公路上从开始到 i 英里产生的最大收入。现在对于高速公路上的每一英里，我们需要检查这一英里是否有任何广告牌的选项，如果没有，那么直到该英里产生的最大收入将与直到一英里之前产生的最大收入相同。但是，如果该英里可以选择广告牌，那么我们有两个选择：
1.要么我们将放置广告牌，忽略前t英里的广告牌，并添加放置的广告牌的收入。
2. 忽略这个广告牌。所以 maxRev[i] = max(maxRev[it-1] + Revenue[i], maxRev[i-1])
下面是这种方法的实现：

C++

// C++ program to find maximum revenue by placing
// billboard on the highway with given constraints.
#include
using namespace std;
 
int maxRevenue(int m, int x[], int revenue[], int n,
                                              int t)
{
    // Array to store maximum revenue at each miles.
    int maxRev[m+1];
    memset(maxRev, 0, sizeof(maxRev));
 
    // actual minimum distance between 2 billboards.
    int nxtbb = 0;
    for (int i = 1; i <= m; i++)
    {
        // check if all billboards are already placed.
        if (nxtbb < n)
        {
            // check if we have billboard for that particular
            // mile. If not, copy the previous maximum revenue.
            if (x[nxtbb] != i)
                maxRev[i] = maxRev[i-1];
 
            // we do have billboard for this mile.
            else
            {
                // We have 2 options, we either take current
                // or we ignore current billboard.
 
                // If current position is less than or equal to
                // t, then we can have only one billboard.
                if (i <= t)
                    maxRev[i] = max(maxRev[i-1], revenue[nxtbb]);
 
                // Else we may have to remove previously placed
                // billboard
                else
                    maxRev[i] = max(maxRev[i-t-1]+revenue[nxtbb],
                                                  maxRev[i-1]);
 
                nxtbb++;
            }
        }
        else
            maxRev[i] = maxRev[i - 1];
    }
 
    return maxRev[m];
}
 
// Driven Program
int main()
{
    int m = 20;
    int x[] = {6, 7, 12, 13, 14};
    int revenue[] = {5, 6, 5, 3, 1};
    int n = sizeof(x)/sizeof(x[0]);
    int t = 5;
    cout << maxRevenue(m, x, revenue, n, t) << endl;
    return 0;
}

Java

// Java program to find maximum revenue
// by placing billboard on the highway
// with given constraints.
 
class GFG
{
     
static int maxRevenue(int m, int[] x,
                    int[] revenue,
                    int n, int t)
{
     
    // Array to store maximum revenue
    // at each miles.
    int[] maxRev = new int[m + 1];
    for(int i = 0; i < m + 1; i++)
        maxRev[i] = 0;
 
    // actual minimum distance between
    // 2 billboards.
    int nxtbb = 0;
    for (int i = 1; i <= m; i++)
    {
        // check if all billboards are
        // already placed.
        if (nxtbb < n)
        {
            // check if we have billboard for
            // that particular mile. If not,
            // copy the previous maximum revenue.
            if (x[nxtbb] != i)
                maxRev[i] = maxRev[i - 1];
 
            // we do have billboard for this mile.
            else
            {
                // We have 2 options, we either take
                // current or we ignore current billboard.
 
                // If current position is less than or
                // equal to t, then we can have only
                // one billboard.
                if (i <= t)
                    maxRev[i] = Math.max(maxRev[i - 1],
                                        revenue[nxtbb]);
 
                // Else we may have to remove
                // previously placed billboard
                else
                    maxRev[i] = Math.max(maxRev[i - t - 1] +
                                        revenue[nxtbb],
                                        maxRev[i - 1]);
 
                nxtbb++;
            }
        }
        else
            maxRev[i] = maxRev[i - 1];
    }
 
    return maxRev[m];
}
 
// Driver Code
public static void main(String []args)
{
    int m = 20;
    int[] x = new int[]{6, 7, 12, 13, 14};
    int[] revenue = new int[]{5, 6, 5, 3, 1};
    int n = x.length;
    int t = 5;
    System.out.println(maxRevenue(m, x, revenue, n, t));
}
}
 
// This code is contributed by Ita_c.

Python3

# Python3 program to find maximum revenue
# by placing billboard on the highway with
# given constraints.
 
def maxRevenue(m, x, revenue, n, t) :
     
    # Array to store maximum revenue
    # at each miles.
    maxRev = [0] * (m + 1)
 
    # actual minimum distance between
    # 2 billboards.
    nxtbb = 0;
    for i in range(1, m + 1) :
         
        # check if all billboards are
        # already placed.
        if (nxtbb < n) :
             
            # check if we have billboard for
            # that particular mile. If not,
            # copy the previous maximum revenue.
            if (x[nxtbb] != i) :
                maxRev[i] = maxRev[i - 1]
 
            # we do have billboard for this mile.
            else :
             
                # We have 2 options, we either take
                # current or we ignore current billboard.
 
                # If current position is less than or
                # equal to t, then we can have only
                # one billboard.
                if (i <= t) :
                    maxRev[i] = max(maxRev[i - 1],
                                    revenue[nxtbb])
 
                # Else we may have to remove
                # previously placed billboard
                else :
                    maxRev[i] = max(maxRev[i - t - 1] +
                                    revenue[nxtbb],
                                    maxRev[i - 1]);
 
                nxtbb += 1
     
        else :
             
            maxRev[i] = maxRev[i - 1]
     
    return maxRev[m]
     
# Driver Code
if __name__ == "__main__" :
     
    m = 20
    x = [6, 7, 12, 13, 14]
    revenue = [5, 6, 5, 3, 1]
    n = len(x)
    t = 5
    print(maxRevenue(m, x, revenue, n, t))
 
# This code is contributed by Ryuga

C#

// C# program to find maximum revenue
// by placing billboard on the highway
// with given constraints.
using System;
 
class GFG
{
static int maxRevenue(int m, int[] x,
                      int[] revenue,
                      int n, int t)
{
    // Array to store maximum revenue
    // at each miles.
    int[] maxRev = new int[m + 1];
    for(int i = 0; i < m + 1; i++)
        maxRev[i] = 0;
 
    // actual minimum distance between
    // 2 billboards.
    int nxtbb = 0;
    for (int i = 1; i <= m; i++)
    {
        // check if all billboards are
        // already placed.
        if (nxtbb < n)
        {
            // check if we have billboard for
            // that particular mile. If not,
            // copy the previous maximum revenue.
            if (x[nxtbb] != i)
                maxRev[i] = maxRev[i - 1];
 
            // we do have billboard for this mile.
            else
            {
                // We have 2 options, we either take
                // current or we ignore current billboard.
 
                // If current position is less than or
                // equal to t, then we can have only
                // one billboard.
                if (i <= t)
                    maxRev[i] = Math.Max(maxRev[i - 1],
                                         revenue[nxtbb]);
 
                // Else we may have to remove
                // previously placed billboard
                else
                    maxRev[i] = Math.Max(maxRev[i - t - 1] +
                                         revenue[nxtbb],
                                         maxRev[i - 1]);
 
                nxtbb++;
            }
        }
        else
            maxRev[i] = maxRev[i - 1];
    }
 
    return maxRev[m];
}
 
// Driver Code
static void Main()
{
    int m = 20;
    int[] x = new int[]{6, 7, 12, 13, 14};
    int[] revenue = new int[]{5, 6, 5, 3, 1};
    int n = x.Length;
    int t = 5;
    Console.Write(maxRevenue(m, x, revenue, n, t));
}
}
 
// This code is contributed by DrRoot_

Javascript

 
// Javascript program to find maximum revenue
// by placing billboard on the highway
// with given constraints.
     
    function maxRevenue(m,x,revenue,n,t)
    {
    // Array to store maximum revenue
    // at each miles.
    let maxRev = new Array(m + 1);
    for(let i = 0; i < m + 1; i++)
        maxRev[i] = 0;
   
    // actual minimum distance between
    // 2 billboards.
    let nxtbb = 0;
    for (let i = 1; i <= m; i++)
    {
        // check if all billboards are
        // already placed.
        if (nxtbb < n)
        {
            // check if we have billboard for
            // that particular mile. If not,
            // copy the previous maximum revenue.
            if (x[nxtbb] != i)
                maxRev[i] = maxRev[i - 1];
   
            // we do have billboard for this mile.
            else
            {
                // We have 2 options, we either take
                // current or we ignore current billboard.
   
                // If current position is less than or
                // equal to t, then we can have only
                // one billboard.
                if (i <= t)
                    maxRev[i] = Math.max(maxRev[i - 1],
                                        revenue[nxtbb]);
   
                // Else we may have to remove
                // previously placed billboard
                else
                    maxRev[i] = Math.max(maxRev[i - t - 1] +
                                        revenue[nxtbb],
                                        maxRev[i - 1]);
   
                nxtbb++;
            }
        }
        else
            maxRev[i] = maxRev[i - 1];
    }
   
    return maxRev[m];
    }
     
    // Driver Code
    let  m = 20;
    let x=[6, 7, 12, 13, 14];
    let revenue=[5, 6, 5, 3, 1];
    let n = x.length;
    let t = 5;
    document.write(maxRevenue(m, x, revenue, n, t));
     
    // This code is contributed by rag2127
 

输出：

10

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