题目描述

Being a fan of all cold-weather sports (especially those involving cows),

Farmer John wants to record as much of the upcoming winter Moolympics as

possible.

The television schedule for the Moolympics consists of N different programs

(1 <= N <= 150), each with a designated starting time and ending time. FJ

has a dual-tuner recorder that can record two programs simultaneously.

Please help him determine the maximum number of programs he can record in

total. 冬奥会的电视时刻表包含N (1 <= N <= 150)个节目，每个节目都有开始和结束时间。农民约翰有两台录像机，请计算他最多可以录制多少个节目。

输入输出格式

输入格式：

• Line 1: The integer N.

• Lines 2..1+N: Each line contains the start and end time of a single

program (integers in the range 0..1,000,000,000).

输出格式：

• Line 1: The maximum number of programs FJ can record.

输入输出样例

6

0 3

6 7

3 10

1 5

2 8

1 9

4

说明

INPUT DETAILS:

The Moolympics broadcast consists of 6 programs. The first runs from time

0 to time 3, and so on.

OUTPUT DETAILS:

FJ can record at most 4 programs. For example, he can record programs 1

and 3 back-to-back on the first tuner, and programs 2 and 4 on the second

tuner.

Source: USACO 2014 January Contest, Silver

分析

如果只有一台的话直接贪心即可，两台呢，贪心也可以，多加一个变量，记录另一台摄像机到什么时间了即可

额能会有几种情况：

• 两台摄像机都在拍摄节目，这个节目也就拍不成了
• 有一台摄像机空闲，那就选这台拍
• 两台都空闲，那么找两者结束之前拍摄较晚的继续拍下去，另外一台留给后面的拍，这样选择余地更多（浪费也少）

注意结束时间还是可以用的，他在结束时间的前一分钟已经拍完了，手动测试样例发现的，难怪我调试不对orz。

代码

 #include<cstdio>

 #include<algorithm>

 using namespace std;

 struct Que{

     int l,r;

     bool operator < (const Que &a) const

     {

         return r < a.r;

     }

 }q[];

 int main()

 {

     int n,ans = , last1 = -,last2 = -;

     scanf("%d",&n);

     for (int i=; i<=n; ++i)

         scanf("%d%d",&q[i].l,&q[i].r);

     sort(q+,q+n+);

     for (int i=; i<=n; ++i)

     {

         if (q[i].l<last1&&q[i].l<last2) continue ;

         ans++;

         if (q[i].l>=last1&&q[i].l<last2) last1 = q[i].r;

         else if (q[i].l<last1&&q[i].r>=last2) last2 = q[i].r;

         else if (last1<last2) last2 = q[i].r;

         else last1 = q[i].r;

     }

     printf("%d",ans);

     return ;

 }