活动选择问题
共有n种不同的活动，分别包含其开始时间和结束时间。选择一个人最多可以解决的活动数。我们将使用贪婪方法找到其余活动中完成时间最少的下一个活动，并且开始时间大于或等于上一个选定活动的结束时间。
当列表未排序时，此问题的复杂度为O（nlogn）。
提供排序列表后，复杂度将为O(n)。
输入输出
Input:
A list of different activities with starting and ending times.
{(5,9), (1,2), (3,4), (0,6), (5,7), (8,9)}
Output:
Selected Activities are:
Activity: 0 , Start: 1 End: 2
Activity: 1 , Start: 3 End: 4
Activity: 3 , Start: 5 End: 7
Activity: 5 , Start: 8 End: 9算法
maxActivity(act, size)
输入：活动列表，以及列表中的元素数。
输出- 选择活动的顺序。
Begin
initially sort the given activity List
set i := 1
display the ith activity //in this case it is the first activity
for j := 1 to n-1 do
if start time of act[j] >= end of act[i] then
display the jth activity
i := j
done
End示例
#include<iostream>
#include<algorithm>
using namespace std;
struct Activitiy {
int start, end;
};
bool comp(Activitiy act1, Activitiy act2) {
return (act1.end < act2.end);
}
void maxActivity(Activitiy act[], int n) {
sort(act, act+n, comp); //sort activities using compare function
cout << "Selected Activities are: " << endl;
int i = 0;// first activity as 0 is selected
cout << "Activity: " << i << " , Start: " <<act[i].start << " End:
" << act[i].end <<endl;
for (int j = 1; j < n; j++) { //for all other activities
if (act[j].start >= act[i].end) { //when start time is >= end
time, print the activity
cout << "Activity: " << j << " , Start: " <<act[j].start << " End: " << act[j].end <<endl;
i = j;
}
}
}
int main() {
Activitiy actArr[] = {{5,9},{1,2},{3,4},{0,6},{5,7},{8,9}};
int n = 6;
maxActivity(actArr,n);
return 0;
}输出结果
Selected Activities are: Activity: 0 , Start: 1 End: 2 Activity: 1 , Start: 3 End: 4 Activity: 3 , Start: 5 End: 7 Activity: 5 , Start: 8 End: 9