码迷,mamicode.com
首页 > 其他好文 > 详细

3433: [Usaco2014 Jan]Recording the Moolympics

时间:2015-04-13 16:23:52      阅读:169      评论:0      收藏:0      [点我收藏+]

标签:

3433: [Usaco2014 Jan]Recording the Moolympics

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 137  Solved: 89
[Submit][Status][Discuss]

Description

 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.

给出n个区间[a,b).有2个记录器.每个记录器中存放的区间不能重叠.

求2个记录器中最多可放多少个区间.

 

 

Input

* 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).

Output

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

Sample Input

6
0 3
6 7
3 10
1 5
2 8
1 9

INPUT DETAILS: The Moolympics broadcast consists of 6 programs. The first runs from time 0 to time 3, and so on.

Sample Output

4

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.

HINT

 

Source

Silver 鸣谢Alegria_提供译文

 

题解:(呵呵哒我会告诉你我的第一反应是网络流?)

其实仔细看看后发现还是个贪心,只不过现在是两个容器= =,然后就是经典的右边界排序后O(n)乱搞了

 1 /**************************************************************
 2     Problem: 3433
 3     User: HansBug
 4     Language: Pascal
 5     Result: Accepted
 6     Time:0 ms
 7     Memory:304 kb
 8 ****************************************************************/
 9  
10 var
11    i,j,k,l,m,n,ans,l1,l2:longint;
12    a:array[0..10000,1..2] of longint;
13 procedure swap(var x,y:longint);
14           var z:longint;
15           begin
16                z:=x;x:=y;y:=z;
17           end;
18 procedure sort(l,r:longint);
19           var i,j,x,y:longint;
20           begin
21                i:=l;j:=r;x:=a[(l+r) div 2,2];
22                repeat
23                      while a[i,2]<x do inc(i);
24                      while a[j,2]>x do dec(j);
25                      if i<=j then
26                         begin
27                              swap(a[i,1],a[j,1]);
28                              swap(a[i,2],a[j,2]);
29                              inc(i);dec(j);
30                         end;
31                until i>j;
32                if i<r then sort(i,r);
33                if l<j then sort(l,j);
34           end;
35  
36 begin
37      readln(n);
38      for i:=1 to n do readln(a[i,1],a[i,2]);
39      sort(1,n);
40      for i:=1 to n do
41          begin
42               if (a[i,1]>=l1) then
43                  begin
44                       l1:=a[i,2];
45                       inc(ans);
46                  end
47               else if (a[i,1]>=l2) then
48                    begin
49                         l2:=a[i,2];
50                         inc(ans);
51                    end;
52               if l1<l2 then swap(l1,l2);
53          end;
54      writeln(ans);
55      readln;
56 end.       

 

3433: [Usaco2014 Jan]Recording the Moolympics

标签:

原文地址:http://www.cnblogs.com/HansBug/p/4422286.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!