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
注意：根据需要在M英里的高速公路上放置广告牌，所有可能的站点从x 1到x n都在0到M范围内。
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
令maxRev [i]，1 <= i <= M，是从高速公路开始到i英里所产生的最大收益。现在，对于高速公路上的每一英里，我们需要检查该英里是否可以选择任何广告牌，否则，直到该英里之前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，如果该英里有广告牌选项，那么我们有2个选项：
1.我们要么放置广告牌，要么忽略前t英里内的广告牌，然后添加所放置广告牌的收入。
2.忽略此广告牌。因此maxRev [i] = max(maxRev [it-1] +收入[i]，maxRev [i-1])
以下是此方法的实现：
C++
// C++ program to find maximum revenue by placing
// billboard on the highway with given constarints.
#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 constarints.
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 constarints.
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 RyugaC#
// C# program to find maximum revenue
// by placing billboard on the highway
// with given constarints.
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_PHP
输出：
10
时间复杂度： O(M)，其中M是总公路的距离。
辅助空间： O(M)。
来源 ：
https://courses.cs.washington.edu/courses/cse421/06au/slides/Lecture18/Lecture18.pdf