一家公司有一间会议室。有(S[i], F[i])形式的N 个会议，其中S[i]是会议i的开始时间， F[i]是会议i 的结束时间。任务是找出会议室可容纳的最大会议数。打印所有会议号
例子：
Input : s[] = {1, 3, 0, 5, 8, 5}, f[] = {2, 4, 6, 7, 9, 9}
Output : 1 2 4 5
First meeting [1, 2]
Second meeting [3, 4]
Fourth meeting [5, 7]
Fifth meeting [8, 9]
Input : s[] = {75250, 50074, 43659, 8931, 11273, 27545, 50879, 77924},
f[] = {112960, 114515, 81825, 93424, 54316, 35533, 73383, 160252 }
Output : 6 7 1
方法：
想法是使用贪心方法解决问题，这与活动选择问题相同。
- 按每对的第二个数字（完成时间）的升序对所有对（会议）进行排序。
- 选择已排序对的第一个会议作为房间中的第一个会议，并将其推入结果向量，并使用第一个选定会议的第二个值（结束时间）设置变量 time_limet(say)。
- 从数组的第二对到最后一对迭代，如果当前对的第一个元素（会议开始时间）的值大于先前选择的对完成时间（time_limit），则选择当前对并更新结果向量（将选定的会议号推入向量）和变量 time_limit。
- 从向量打印会议顺序。
下面是上述方法的实现。
C++
// C++ program to find maximum number of meetings
#include
using namespace std;
// Structure for storing starting time,
// finishing time and position of meeting.
struct meeting {
int start;
int end;
int pos;
};
// Comparator function which can compare
// the second element of structure used to
// sort pairs in increasing order of second value.
bool comparator(struct meeting m1, meeting m2)
{
return (m1.end < m2.end);
}
// Function for finding maximum meeting in one room
void maxMeeting(int s[], int f[], int n)
{
struct meeting meet[n];
for (int i = 0; i < n; i++)
{
// Starting time of meeting i.
meet[i].start = s[i];
// Finishing time of meeting i
meet[i].end = f[i];
// Original position/index of meeting
meet[i].pos = i + 1;
}
// Sorting of meeting according to their finish time.
sort(meet, meet + n, comparator);
// Vector for storing selected meeting.
vector m;
// Initially select first meeting.
m.push_back(meet[0].pos);
// time_limit to check whether new
// meeting can be conducted or not.
int time_limit = meet[0].end;
// Check for all meeting whether it
// can be selected or not.
for (int i = 1; i < n; i++) {
if (meet[i].start >= time_limit)
{
// Push selected meeting to vector
m.push_back(meet[i].pos);
// Update time limit.
time_limit = meet[i].end;
}
}
// Print final selected meetings.
for (int i = 0; i < m.size(); i++) {
cout << m[i] << " ";
}
}
// Driver code
int main()
{
// Starting time
int s[] = { 1, 3, 0, 5, 8, 5 };
// Finish time
int f[] = { 2, 4, 6, 7, 9, 9 };
// Number of meetings.
int n = sizeof(s) / sizeof(s[0]);
// Function call
maxMeeting(s, f, n);
return 0;
} Java
// Java program to find maximum number of meetings
import java.util.*;
// Comparator function which can compare
// the ending time of the meeting ans
// sort the list
class mycomparator implements Comparator
{
@Override
public int compare(meeting o1, meeting o2)
{
if (o1.end < o2.end)
{
// Return -1 if second object is
// bigger then first
return -1;
}
else if (o1.end > o2.end)
// Return 1 if second object is
// smaller then first
return 1;
return 0;
}
}
// Custom class for storing starting time,
// finishing time and position of meeting.
class meeting
{
int start;
int end;
int pos;
meeting(int start, int end, int pos)
{
this.start = start;
this.end = end;
this.pos = pos;
}
}
class GFG{
// Function for finding maximum meeting in one room
public static void maxMeeting(ArrayList al, int s)
{
// Initialising an arraylist for storing answer
ArrayList m = new ArrayList<>();
int time_limit = 0;
mycomparator mc = new mycomparator();
// Sorting of meeting according to
// their finish time.
Collections.sort(al, mc);
// Initially select first meeting.
m.add(al.get(0).pos);
// time_limit to check whether new
// meeting can be conducted or not.
time_limit = al.get(0).end;
// Check for all meeting whether it
// can be selected or not.
for(int i = 1; i < al.size(); i++)
{
if (al.get(i).start > time_limit)
{
// Add selected meeting to arraylist
m.add(al.get(i).pos);
// Update time limit
time_limit = al.get(i).end;
}
}
// Print final selected meetings.
for(int i = 0; i < m.size(); i++)
System.out.print(m.get(i) + 1 + " ");
}
// Driver Code
public static void main (String[] args)
{
// Starting time
int s[] = { 1, 3, 0, 5, 8, 5 };
// Finish time
int f[] = { 2, 4, 6, 7, 9, 9 };
// Defining an arraylist of meet type
ArrayList meet = new ArrayList<>();
for(int i = 0; i < s.length; i++)
// Creating object of meeting
// and adding in the list
meet.add(new meeting(s[i], f[i], i));
// Function call
maxMeeting(meet, meet.size());
}
}
// This code is contributed by Sarthak Sethi Python3
# Python3 program to find maximum number
# of meetings
# Custom class for storing starting time,
# finishing time and position of meeting.
class meeting:
def __init__(self, start, end, pos):
self.start = start
self.end = end
self.pos = pos
# Function for finding maximum
# meeting in one room
def maxMeeting(l, n):
# Initialising an arraylist
# for storing answer
ans = []
# Sorting of meeting according to
# their finish time.
l.sort(key = lambda x: x.end)
# Initially select first meeting
ans.append(l[0].pos)
# time_limit to check whether new
# meeting can be conducted or not.
time_limit = l[0].end
# Check for all meeting whether it
# can be selected or not.
for i in range(1, n):
if l[i].start > time_limit:
ans.append(l[i].pos)
time_limit = l[i].end
# Print final selected meetings
for i in ans:
print(i + 1, end = " ")
print()
# Driver code
if __name__ == '__main__':
# Starting time
s = [ 1, 3, 0, 5, 8, 5 ]
# Finish time
f = [ 2, 4, 6, 7, 9, 9 ]
# Number of meetings.
n = len(s)
l = []
for i in range(n):
# Creating object of meeting
# and adding in the list
l.append(meeting(s[i], f[i], i))
# Function call
maxMeeting(l, n)
# This code is contributed by MuskanKalra1输出：
1 2 4 5时间复杂度： O(N log(N))
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