Ytu 2892 free Watch movie greedy

2892:i--free to watch moviesTime limit:1 Sec Memory limit:128 MB
Submit:22 solved:10
[Submit] [Status] [Web Board] Description

Vientiane City Star U.S. cinema opening 1 years, to hold a one-day big bargain, the ACM players ready to go to the movies. Known cinemas play at various stages of the movie. Ask the ACM team how to arrange their own time, so that the number of movies to see the most. Input

Test data first behavior N (n>=0), which represents the number of movies, followed by n rows, each row has two integers, indicating the start and end time of each movie. Output

Output the maximum number of films that the ACM can watch. Sample Input

A
8 of
16 20 in a-one-and-a-list

Sample Output

5

HINT

Source

LDF

AC Code:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;
struct movie{
    int bg;
    int en;
};
Movie my[111];
BOOL Comp (movie A,movie b) {
    if (a.bg!=b.bg) return a.en<b.en;
    Return a.bg<b.bg;
}
int main () {
    int n;
    cin>>n;
    int i;
    for (i=0;i<n;++i) {
        cin>>my[i].bg>>my[i].en;
    }
    Sort (my,my+n,comp);
    /**
    cout<< ' \ n ';
    for (i=0;i<n;++i) cout<<my[i].bg<< ' <<my[i].en<< ' \12 ';
    cout<< ' \ n ';
    /**/
    int sum=1,t;
    T=my[0].en;
    for (i=0;i<n;++i) {
        if (t<=my[i].bg) {
            sum++;
            T=my[i].en
        }
    }
    if (n==0) sum=0;
    cout<<sum<< ' \12 ';
    return 0;
}

LDF