活动选择问题的Java程序|贪心算法-1
您有 n 个活动及其开始和结束时间。选择一个人可以执行的最大活动数,假设一个人一次只能从事一项活动。
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[] ]
Java
// The following implementation assumes that the activities
// are already sorted according to their finish time
import java.util.*;
import java.lang.*;
import java.io.*;
class ActivitySelection {
// 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
public static void printMaxActivities(int s[], int f[], int n)
{
int i, j;
System.out.print("Following activities are selected : n");
// The first activity always gets selected
i = 0;
System.out.print(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]) {
System.out.print(j + " ");
i = j;
}
}
}
// driver program to test above function
public static void main(String[] args)
{
int s[] = { 1, 3, 0, 5, 8, 5 };
int f[] = { 2, 4, 6, 7, 9, 9 };
int n = s.length;
printMaxActivities(s, f, n);
}
}
输出:
Following activities are selected : n0 1 3 4
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