题目描述
[题目描述] 在大公司里，会议是很多的，开会得有场子，要场子你得先在电子流里预订。 [Input] [Output] [I/O Example] Output |
练习模板
#include <cstdio>
#include <iostream>
using
namespace std;
int main(
int argc,
char** argv)
{
int tc, T;
cin >> T;
for(tc =
0; tc < T; tc++)
{
//
TO DO here
}
return
0;
//
Your program should return 0 on normal termination.
}
代码实现
题目中要求会议时间不可以冲突，所以可以利用贪心算法，尽可能的选择会议时间结束较早的会议室，这样就能安排最多的会议室。
#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
using
namespace std;
class MeetingRoom {
public:
int index;
int start;
int end;
public:
MeetingRoom(
int i,
int s,
int e) : index(i), start(s), end(e) {}
};
bool isEndEarly(
const MeetingRoom r1,
const MeetingRoom r2) {
return (r1.end < r2.end);
}
int main(
int argc,
char** argv)
{
int tc, T;
int num =
0;
int count =
0;
int index =
0;
int start =
0;
int end =
0;
vector<MeetingRoom> rooms;
freopen(
“
input.txt
“,
“
r
“, stdin);
cin >> T;
for(tc =
0; tc < T; tc++)
{
rooms.clear();
//
clear vector
//
get all the info of rooms
cin >> num;
for (
int i =
0; i < num; i++) {
cin >> index >> start >> end;
MeetingRoom room(index, start, end);
rooms.push_back(room);
}
//
sort rooms for end time
sort(rooms.begin(), rooms.end(), isEndEarly);
//
use greedy algorithm to solve promblem
count =
1;
MeetingRoom prev = rooms[
0];
for (vector<MeetingRoom>::iterator it = rooms.begin() +
1; it != rooms.end(); it++) {
MeetingRoom current = *it;
if (current.start >= prev.end) {
count++;
prev = current;
}
}
cout << count << endl;
}
return
0;
}
