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// C++ program for activity selection problem. | ||
// The following implementation assumes that the activities | ||
// are already sorted according to their finish time | ||
#include <bits/stdc++.h> | ||
using namespace std; | ||
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// 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; | ||
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cout <<"Following activities are selected "<< endl; | ||
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// The first activity always gets selected | ||
i = 0; | ||
cout <<" "<< i; | ||
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// 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]) | ||
{ | ||
cout <<" " << j; | ||
i = j; | ||
} | ||
} | ||
} | ||
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// 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; | ||
} | ||
//this code contributed by shivanisinghss2110 |