NYOJ 14 venue arrangement problem (also a classic problem)

Venue schedule problem time limit: 3000 MS | memory limit: 65535 KB difficulty: 4

Description

There are many activities in the auditorium of the school every day. The scheduled time for these activities may conflict, and you need to select some activities to hold. Xiao Liu's job is to arrange activities in the school auditorium. At most one activity can be arranged at a time. Now Mr. Liu has some schedules for his activities. He wants to arrange as many activities as possible. How can he arrange them.

Input

The first row is an integer m (m <100), indicating a total of m groups of test data.
The first row of test data in each group is an integer n (1 <n <10000), indicating that the test data has n activities.
Next n rows, each row has two positive integers Bi, Ei (0 <= Bi, Ei <10000), indicating the start time and end time of the I-th activity (Bi <= Ei)

Output

For each group of inputs, the maximum number of activities that can be arranged is output.
Output of each group occupies one row

Sample input

221 1010 1131 1010 1111 20

Sample output

12

Prompt

Note: If the last activity ends at t time, the next activity should start at t + 1 at the earliest.

Algorithm analysis:

There are three kinds of algorithms that can be considered for scheduling without overlapping intervals:

1. In the optional work, the earliest end time is selected each time.

2. In the optional work, the shortest job is selected every time.

3. In an optional job, work that overlaps with the minimum number of optional jobs is selected each time.

Facts have proved that only one of them is correct and can withstand the test.

For the first algorithm, the final end time of the algorithm when the same number of jobs started earlier is not later than that of other schemes.

That is to say, the number after the first ending interval must be greater than or equal to the number of other later intervals.

For details, see: http://blog.csdn.net/luoweifu/article/details/18195607

The code is as follows:

# Include <iostream> # include <utility> # include <algorithm> using namespace std; typedef pair <int, int> name; name arr [10005]; int cmp (name, name B) {return. second <B. second;} int main () {int N, n, I, ans, t; cin> N; while (N --) {cin> n; for (I = 0; I <n; I ++) cin> arr [I]. first> arr [I]. second; sort (arr, arr + n, cmp); ans = 0, t =-1; for (I = 0; I <n; I ++) {if (t <arr [I]. first) {ans ++; t = arr [I]. second ;}} cout <ans <endl ;}return 0 ;}

Because sort sorts the pair type first by default .. I realized it only after WA.

NYOJ 14 venue arrangement problem (also a classic problem)