活动选择问题的C程序|贪婪算法1

您将获得n个活动的开始和结束时间。假设一个人一次只能从事一项活动，请选择一个人可以执行的最大活动数。
例子：

Example 1 : Consider the following 3 activities sorted by finish time.
     start[]  =  {10, 12, 20};
     finish[] =  {20, 25, 30};
A person can perform at most two activities. The 
maximum set of activities that can be executed 
is {0, 2} [ These are indexes in start[] and 
finish[] ]

Example 2 : Consider the following 6 activities 
sorted by by finish time.
     start[]  =  {1, 3, 0, 5, 8, 5};
     finish[] =  {2, 4, 6, 7, 9, 9};
A person can perform at most four activities. The 
maximum set of activities that can be executed 
is {0, 1, 3, 4} [ These are indexes in start[] and 
finish[] ]

C++

// C++ program for activity selection problem.
// The following implementation assumes that the activities
// are already sorted according to their finish time
#include 
  
// Prints a maximum set of activities that can be done by a single
// person, one at a time.
// n   -->  Total number of activities
// s[] -->  An array that contains start time of all activities
// f[] -->  An array that contains finish time of all activities
void printMaxActivities(int s[], int f[], int n)
{
    int i, j;
  
    printf("Following activities are selected n");
  
    // The first activity always gets selected
    i = 0;
    printf("%d ", i);
  
    // Consider rest of the activities
    for (j = 1; j < n; j++) {
        // If this activity has start time greater than or
        // equal to the finish time of previously selected
        // activity, then select it
        if (s[j] >= f[i]) {
            printf("%d ", j);
            i = j;
        }
    }
}
  
// driver program to test above function
int main()
{
    int s[] = { 1, 3, 0, 5, 8, 5 };
    int f[] = { 2, 4, 6, 7, 9, 9 };
    int n = sizeof(s) / sizeof(s[0]);
    printMaxActivities(s, f, n);
    return 0;
}

输出：

Following activities are selected n0 1 3 4

请参阅有关活动选择问题的完整文章。贪婪的Algo-1了解更多详情！

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