


C++ program to calculate how many movies at a film festival participants can watch in full
Suppose there is a film festival that showcases various films from different countries. Now, a participant wants to attend as many non-overlapping movies as possible, and we need to help them figure out how many movies they can attend.
There is a structure Movie, which has the following members:
- The starting time of the movie.
- The duration of the movie.
- The end time of the movie.
There is also a structure Festival, which has the following members:
- The number of films in the festival.
- An array of type Movie whose size is the same as the number of movies in the festival.
We need to create and initialize a Festival object, which contains two arrays 'timeBegin' and 'duration', which contain the start time and duration of multiple movies respectively. The integer n represents the total number of movies and is also used to initialize the object. We further use this object to calculate how many movies a participant can watch in full.
So if the input is timeBegin = {1, 3, 0, 5, 5, 8, 8}, duration = {3, 2, 2, 4, 3, 2, 3}, n = 7 , then the output will be 4
Participants can watch 4 movies in full at the festival.
In order to solve this problem, we will follow the following steps:
- Structure Movie {
- Define three member variables timeBegin, duration, timeEnd
- Overloaded operator '
- Return timeEnd
- Define a member variable count
- Define an array containing Movie type items movies
- filmFestival := a new Festival object
- filmFestival’s count := count
- For initialization i := 0, when i
- temp := timeBegin of a new Movie type object
- temp := timeBegin[i]
- temp duration := duration[ i]
- timeEnd of temp := timeBegin[i] duration[i]
- Insert temp into the movies array of filmFestival
- res := 0
- Movies for fest The array is sorted
- timeEnd := -1
- For initialization i := 0, when i count, update (increase i by 1), perform the following operations:
- If timeBegin of fest's movies[i] >= timeEnd, perform the following operations:
- (Increase res by 1)
- timeEnd := timeEnd of fest's movies[i]
- If timeBegin of fest's movies[i] >= timeEnd, perform the following operations:
- Return res
Example
Let us look at the following implementation to better Well understood:
#include<bits/stdc++.h> using namespace std; struct Movie { int timeBegin, duration, timeEnd; bool operator<(const Movie& another) const { return timeEnd < another.timeEnd; } }; struct Festival { int count; vector<Movie> movies; }; Festival* initialize(int timeBegin[], int duration[], int count) { Festival* filmFestival = new Festival; filmFestival->count = count; for (int i = 0; i < count; i++) { Movie temp; temp.timeBegin = timeBegin[i]; temp.duration = duration[i]; temp.timeEnd = timeBegin[i] + duration[i]; filmFestival->movies.push_back(temp); } return filmFestival; } int solve(Festival* fest) { int res = 0; sort(fest->movies.begin(), fest->movies.end()); int timeEnd = -1; for (int i = 0; i < fest->count; i++) { if (fest->movies[i].timeBegin >= timeEnd) { res++; timeEnd = fest->movies[i].timeEnd; } } return res; } int main(int argc, char *argv[]) { int timeBegin[] = {1, 3, 0, 5, 5, 8, 8}; int duration[] = {3, 2, 2, 4, 3, 2, 3}; Festival * fest; fest = initialize(timeBegin,duration, 7); cout << solve(fest) << endl; return 0; }
input
int timeBegin[] = {1, 3, 0, 5, 5, 8, 8}; int duration[] = {3, 2, 2, 4, 3, 2, 3}; Festival * fest; fest = initialize(timeBegin,duration, 7);
output
4
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